# Physics (Grades 9–12)

## Subtest 2 Sample Items

1. The diagram below shows a circuit consisting of a switch *S* and several resistors of

resistance *R*.

Which of the following statements best describes what will happen if switch *S* is opened?

The diagram below shows a circuit consisting of a switch S and several resistors of resistance R.
A schematic of a circuit is shown. The upper left hand corner of the circuit is labeled point a. Starting at point a and traversing the circuit from left to right one encounters a single resistor. Next, one encounters two resistors in parallel. Of the two parallel resistors, the top resistor is in series with a switch labeled S. Continuing to the right along a single wire, one encounters the upper right hand corner, which is labeled point b. Continuing to traverse the circuit one encounters a battery before returning to point a.

- The equivalent resistance of the circuit will decrease.
- The power supplied to the circuit will decrease.
- The current through the circuit will increase.
- The voltage across
*a* and *b* will increase.

2. A resistor connected across a constant voltage source dissipates power *P*. If the value of the resistance is doubled and the voltage remains the same, what is the power dissipated by the new resistor?

- 2
*P*
- 4
*P*

- Answer
**Correct Response: B.** (Objective 0010) This question requires the examinee to demonstrate knowledge of measuring and calculating time-varying or constant values of current, voltage, and power in electric circuits. The power consumed by a device in an electric circuit is the product of the voltage and the current, or *P* = *IV*. For a resistor, *V* = *IR*, where *R* is the resistance (Ohm's law). Solving Ohm's law for current gives and by substitution, . Since the voltage remains constant, doubling the resistance will reduce the amount of power dissipated by Joule heating by a factor of one-half . This is due to the fact that increasing the resistance will decrease the current, and therefore the power consumed.

Correct Response: B. (Objective 0010) This question requires the examinee to demonstrate knowledge of measuring and calculating time-varying or constant values of current, voltage, and power in electric circuits. The power consumed by a device in an electric circuit is the product of the voltage and the current, or P equals I V. For a resistor, V equals I R, where R is the resistance (Ohm's law). Solving Ohm's law for current gives I equals V over R and by substitution, P equals V squared over R. Since the voltage remains constant, doubling the resistance will reduce the amount of power dissipated by Joule heating by a factor of one-half, open parenthesis one-half P close parenthesis. This is due to the fact that increasing the resistance will decrease the current, and therefore the power consumed.

3. All of the resistors in the circuit below are identical and the battery is ideal.

The circuit diagram indicates a voltage source in series with a resistor and in series with two resistors that are in parallel. One of the two parallel resistors is in series with a switch.

Each resistor in the circuit has a value of 10 Ω. While the switch is open, a voltmeter placed across the resistor marked *R* reads 40 V. What is the current through the battery when the switch is closed?

- 1.3 A
- 2.6 A
- 3.9 A
- 5.3 A

- Answer
**Correct Response: D.** (Objective 0010) This question requires the examinee to apply Kirchhoff's laws to explain and predict the current through or the voltage across elements in a given circuit. When the switch is open, the circuit consists of two 10 Ω resistors in series, so each of the two resistors has the same current through it. Because the voltage across resistor *R* is 40 V, by *V* = *IR* the voltage across the other resistor is 40 V. By Kirchhoff's voltage law, 40 V + 40 V − *V*_{battery} = 0.0 V, so the voltage across the battery is 80 V. When the switch is closed, the circuit consists of one resistor in series with a parallel circuit consisting of two resistors. The total resistance of the closed circuit is *R*_{T} = 10 Ω + *R*_{eq}, where or *R*_{eq} = 5 Ω, and
*R*_{T} = 15 Ω. This circuit can be replaced with an equivalent circuit of resistance 15 Ω connected in series with an 80 V battery. From *V* = *IR* the current through the battery is 5.3 A.

Correct Response: D. (Objective 0010) This question requires the examinee to apply Kirchhoff's laws to explain and predict the current through or the voltage across elements in a given circuit. When the switch is open, the circuit consists of two 10 Omega resistors in series, so each of the two resistors has the same current through it. Because the voltage across resistor *R* is 40 V, by *V* equals *I R* the voltage across the other resistor is 40 V. By Kirchhoff's voltage law, 40 V plus 40 V minus *V* sub battery equals 0.0 V, so the voltage across the battery is 80 V. When the switch is closed, the circuit consists of one resistor in series with a parallel circuit consisting of two resistors. The total resistance of the closed circuit is *R *sub t equals 10 Omega plus *R* sub e q where one over r sub e q equals one over ten ohms plus one over ten ohms or *R* sub eq equals 5 Omega, and *R* sub t equals 15 Omega. This circuit can be replaced with an equivalent circuit of resistance 15 Omega connected in series with an 80 V battery. From *V* equals *I R* the current through the battery is 5.3 A.

4. A physics student has two 10 Ω resistors and two 20 pF capacitors. The student would like to build a circuit component that requires a time constant of τ = *RC* = 800 ps. Which of the following arrangements could be used?

A. an arrangement of four circuit components in series. From left to right, they are a resistor, a second resistor, a capacitor, and a second capacitor.

B. an arrangement of four circuit components. From left to right, they are two resistors in parallel which are in turn in series with a capacitor and a second capacitor.

C. an arrangement of four circuit components. From left to right, they are a resistor in series with a second resistor, which are in turn in series with two capacitors in parallel with one another.

D. an arrangement of four circuit components. From left to right, they are two resistors in parallel which are in series with two capacitors which are in parallel.

- Answer
**Correct Response: A.** (Objective 0010) This question requires the examinee to demonstrate knowledge of how to design a circuit that behaves in a given way. The time constant for a circuit consisting of capacitors and resistors is the product of the equivalent resistance and capacitance of the circuit. Answering the question requires determining the equivalent resistance and capacitance of each of the circuits. Analyzing each circuit requires using the formulas for two resistors and capacitors in series and parallel: *R*_{s} = *R*_{1} + *R*_{2}, for resistors and , *C*_{p} = *C*_{1} + *C*_{2} for capacitors. Applying these formulas to the circuit consisting of two resistors in series and two capacitors in parallel gives *R*_{s} = *R*_{1} + *R*_{2} = 10 ω + 10 ω = 20 ω and *C*_{p} = *C*_{1} + *C*_{2} = 20 pF + 20 pF = 40 pF. The time constant is τ = *RC* = (20 ω )(40 pF) = 800 ps.

Correct Response: A. (Objective 0010) This question requires the examinee to demonstrate knowledge of how to design a circuit that behaves in a given way. The time constant for a circuit consisting of capacitors and resistors is the product of the equivalent resistance and capacitance of the circuit. Answering the question requires determining the equivalent resistance and capacitance of each of the circuits. Analyzing each circuit requires using the formulas for two resistors and capacitors in series and parallel: *R* sub s equals *R* sub one plus *R* sub two, one over r sub p equals one over r sub one baseline plus one over r sub two for resistors and one over c sub s equals one over c sub one baseline plus one over c sub two, *C* sub p equals *C* sub one plus *C* sub two for capacitors. Applying these formulas to the circuit consisting of two resistors in series and two capacitors in parallel gives *R *sub s equals *R* sub one plus *R* sub two equals 10 omega plus 10 omega equals 20 omega and *C* sub p equals *C* sub one plus *C* sub two equals 20 p F plus 20 p F equals 40 p F. The time constant is tau equals *R C* equals open parenthesis 20 omega close parenthesis open parenthesis 40 p F close parenthesis equals 800 p s.

5. A very large plane surface carries a uniform surface charge density of σ. A cylinder of height *h* and radius *r* is placed on the plane such that the sides of the cylinder are perpendicular to the plane and the top and bottom of the cylinder are parallel to the plane, as shown in the diagram below.

What is the magnitude of the electric field, *E*, through the cylinder above the plane?

A very large plane surface carries a uniform surface charge density of sigma. A cylinder of height h and radius r is placed on the plane such that the sides of the cylinder are perpendicular to the plane and the top and bottom of the cylinder are parallel to the plane, as shown in the diagram below.
The diagram shows a perspective diagram of a plane surface that has plus symbols marked on the surface at regular intervals. Several electric field vectors are drawn perpendicular to the surface. Some of the vectors point upwards from the top of the plane, and other field vectors point downwards from the bottom side of the plane. The plane intersects a right circular cylinder. The sides of the cylinder are perpendicular to the plane. The top and bottom of the cylinder are parallel to the plane. The intersection of the plane and the cylinder is shaded. One electric field vector passes through the top of the cylinder, and one passes through the bottom of the cylinder.
What is the magnitude of the electric field, E, through the cylinder above the plane?

A. E equals sigma over two times epsilon zero.

B. E equals sigma over epsilon zero.

C. E equals sigma over the product of two, epsilon zero, pi, and r.

D. E equals sigma over the product of epsilon zero, pi, and h.

*E* =
*E* =
*E* =
*E* =

6. A particle with positive charge *q* and speed *v*_{0} in the horizontal direction enters a region between the parallel plates of a capacitor. The distance between the plates is *d* and the voltage across the plates is *V*. In addition to a uniform electric field, there is a uniform magnetic field of magnitude *B* between the plates into the page as shown below.

two parallel plates separated by a distance, d and showing a voltage V. Between the two plates field lines of magnitude B are shown moving into the plane of the page. The positive particle labeled with v sub zero and positive q travels into the field from the left moving to the right .

If the net Lorentz force on the particle is zero, what is the speed of the particle?

A. v sub zero equals start fraction numerator v denominator b d

B. v sub zero equals start fraction numerator v d denominator b

C. v sub zero equals start fraction numerator v b denominator d

D. v sub zero equals v b d

- Answer
**Correct Response: A.** (Objective 0011) This question requires the examinee to apply Newton's laws and the Lorentz force to explain and predict the motion of charges in given electric and magnetic fields. The Lorentz force describes the force on a point charge in an electromagnetic field and is given by *F* = *q*(*E* + *v*_{0} × *B*). It is given that the voltage between the plates is *V* and the electric field is uniform. Since the voltage is the work per unit charge, then *V* = *Ed* and the magnitude of the electric field is . The magnetic field is *F* = *qv*_{0} × *B*. *B* and *v*_{0} are perpendicular. Because *B* is into the page, application of the right-hand rule shows that the magnetic force on the particle points from the bottom plate to the top plate of the capacitor. Since the net force on the particle is zero, the electric field must point from the top plate to the bottom plate of the capacitor and the magnitude of the electric force must equal the magnitude of the magnetic force. The magnitude of the magnetic force is *FB* = *qv*_{0}*B* and the magnitude of the electric force is . Setting these two forces equal and cancelling and solving for *v*_{0} gives the result of .

Correct Response: A. (Objective 0011) This question requires the examinee to apply Newton's laws and the Lorentz force to explain and predict the motion of charges in given electric and magnetic fields. The Lorentz force describes the force on a point charge in an electromagnetic field and is given by F equals q open parenthesis E plus v sub zero times B close parenthesis. It is given that the voltage between the plates is V and the electric field is uniform. Since the voltage is the work per unit charge, then V equals E d and the magnitude of the electric field is e equals v over d. The magnetic field is F equals q v sub zero times B. B and v sub zero are perpendicular. Because B is into the page, application of the right-hand rule shows that the magnetic force on the particle points from the bottom plate to the top plate of the capacitor. Since the net force on the particle is zero, the electric field must point from the top plate to the bottom plate of the capacitor and the magnitude of the electric force must equal the magnitude of the magnetic force. The magnitude of the magnetic force is F B equals q v sub zero baseline B and the magnitude of the electric force is q e equals start fraction numerator q v denominator d end fraction. Setting these two forces equal open parenthesis q v sub zero baseline b equals start fraction numerator q v denominator d end fraction and cancelling and solving for v sub zero gives the result of v sub zero equals start fraction numerator v denominator b d end fraction.

7. Three identical point charges of magnitude *q* are located on the vertices of a right triangle with the dimensions shown below.

Right triangle. All vertices are labeled q. The horizontal base or leg is labeled two R and the vertical leg is labeled R. The vertex that is created with the hypotenuse and the long leg that is two R is additionally labeled Y.

What is the magnitude of the horizontal component of the force on the charge located at vertex *Y*?

- Answer
**Correct Response: D.** (Objective 0011) This question requires the examinee to apply Coulomb's law to analyze the forces between point charges. To find the horizontal component of the force on the charge located at vertex *Y*, find the sum of the horizontal components of the forces on the charge from the two other charges. From Coulomb's law, the magnitude of the force on the charge at *Y* from the top charge is , where *r* is the hypotenuse of the triangle so that . The direction of the force vector is along the line shown in the diagram below.
Notice that , so θ = 27°. The horizontal component from this force is thus . The force from the bottom charge points in the horizontal direction and has magnitude . The magnitude of the horizontal component of the force on the charge located at *Y* is the sum of these two quantities: .

Correct Response: D. (Objective 0011) This question requires the examinee to apply Coulomb's law to analyze the forces between point charges. To find the horizontal component of the force on the charge located at vertex Y, find the sum of the horizontal components of the forces on the charge from the two other charges. From Coulomb's law, the magnitude of the force on the charge at Y from the top charge is f sub t equals start fraction numerator k q q denominator r squared end fraction, where r is the hypotenuse of the triangle so that f sub t equals start fraction numerator k q q denominator five r squared end fraction. The direction of the force vector is along the line shown in the following diagram. Right triangle. All vertices are labeled q. The horizontal base or leg is labeled two R and the vertical leg is labeled R. The vertex that is created with the hypotenuse and the long leg that is two R is additionally labeled Y and the angle of the triangle is labeled theta. At point y there is a dash vertical line and a dash horizontal line drawn. the hypotenuse of the triangle is extended past y and the angle created by this extended line and the horizontal dashed line is also labeled theta and f sub t baseline cosine of twenty seven degrees. Notice that tangent of theta equals r over two r equals one half, so theta equals 27 degrees. The horizontal component from this force is thus f sub t baseline cosine twenty seven degrees equals start fraction numerator k q q denominator five r squared cosine of twenty seven degrees. The force from the bottom charge points in the horizontal direction and has magnitude f sub b equals start fraction numerator k q q denominator open parenthesis two r close parenthesis squared end fraction equals start fraction numerator k q q denominator open parenthesis two r close parenthesis squared equals start fraction numerator k q q denominator four r squared end fraction. The magnitude of the horizontal component of the force on the charge located at Y is the sum of these two quantities: start fraction numerator k q squared denominator four r squared end fraction plus start fraction numerator k q squared denominator five p square end fraction cosine twenty seven.

8. The diagram below shows an air core toroid consisting of a wire carrying current I coiled with N loops bent in a circle of radius R.

donut shaped object with radius r. there is a wire labeled i that is coiled around the donut with n number of loops coming onto the donut like object and coming off the object

Assuming that the magnetic field, *B*, is constant in magnitude along and tangent to the dashed circular path inside the toroid, which of the following equations could be used to find the magnetic field?

- 2π
*rB* = μ_{0}*NI*
- 2π
*r* = μ_{0}*NIB*
- π
*r*^{2}B = μ_{0}*NI*
- π
*r*^{2} = μ_{0}*NIB*

- Answer
**Correct Response: A.** (Objective 0011) This question requires the examinee to apply Ampere's law to explain and predict the magnetic field around a given simple geometric distribution of current-carrying wires. The object shown is a toroid, a doughnut-shaped object with wire coiled around its exterior. Current in the wire produces a magnetic field. Assuming that the field is constant along and tangent to the circular curve, Ampere's law can be applied to the dotted circular curve inside the toroid. Ampere's law states that • *d*ℓ = μ_{0}*I*_{enclosed}, where *B* is the magnetic field, • *d*ℓ is the line integral around the closed path *C*, μ_{0} is the magnetic permeability of free space, and Ienclosed is equal to the total current enclosed by the curve *C*. Since the field is constant and tangent to the curve, **B** • *d* ℓ = *Bd* ℓ , and the line integral is just the product of the magnitude of the field **B** and 2π*r*, the length of the curve. The line integral is equal to μ_{0} times the current enclosed by the curve, which can be defined by the current passing through any open surface connecting *C*. If the surface is taken to be a disk centered at *O* with circumference given by the dotted circular curve, then the enclosed current is the number of current loops, *N*, times the current in each loop, *I*. Therefore, Ampere's law gives • *d*ℓ = μ_{0}*I*_{enclosed} = 2π*rB* = μ _{0}*NI*.

Correct Response: A. (Objective 0011) This question requires the examinee to apply Ampere's law to explain and predict the magnetic field around a given simple geometric distribution of current-carrying wires. The object shown is a toroid, a doughnut-shaped object with wire coiled around its exterior. Current in the wire produces a magnetic field. Assuming that the field is constant along and tangent to the circular curve, Ampere's law can be applied to the dotted circular curve inside the toroid. Ampere's law states that closed line integral around the closed curve c of upper bold b dot d l equals mu sub 0 baseline i sub enclosed, where B is the magnetic field, closed line integral around the closed curve c of upper bold b dot d l is the line integral around the closed path C, mu sub 0 is the magnetic permeability of free space, and I sub enclosed is equal to the total current enclosed by the curve C. Since the field is constant and tangent to the curve, B dot d l equals B d l, and the line integral is just the product of the magnitude of the field B and 2 pi r, the length of the curve. The line integral is equal to mu sub 0 times the current enclosed by the curve, which can be defined by the current passing through any open surface connecting C. If the surface is taken to be a disk centered at O with circumference given by the dotted circular curve, then the enclosed current is the number of current loops, N, times the current in each loop, I. Therefore, Ampere's law gives closed line integral around the closed curve c of upper bold b dot d l equals mu sub 0 baseline I sub enclosed equals 2 pi r B equals mu sub 0 baseline N I.

9. A square wire loop of side length *x* is rotated in a constant magnetic field at an angular speed ω, as indicated in the diagram below.

two magnets are shown with field lines traveling from left to right. The square wire loop is in the field lines and shown rotating counterclockwise. The wire loop is connected to a voltmeter with a resistor attached, both of which are outside of the magnetic field.

As the loop rotates, a voltmeter reads a root-mean-square (RMS) value of 80 V across the resistor. If the side of the loop is changed to and the rotational speed of the loop is changed to , what RMS value will the voltmeter measure?

- 10 V
- 20 V
- 40 V
- 80 V

- Answer
**Correct Response: A.** (Objective 0012) This question requires the examinee to apply Lenz's law and Faraday's law to explain and predict induced currents from a given changing magnetic field. As the wire coil rotates, the magnetic field through the coil changes, and by Faraday's law an emf is created in the loop. The voltmeter reads this emf. Faraday's law states that the emf depends on the rate of change of flux through the surface of the coil. The flux is , where **B** is the magnetic field, *A* = *x*^{2} is the area of the loop, and θ = ω*t* is the angle between the vector normal to the surface of the loop and **B**. Using Faraday's law to get the emf, ε , gives . Reducing a side of the square loop to reduces the area to , so the maximum flux and the maximum emf will be reduced to of their previous values from this factor. Slowing the rotational speed, ω , to half its initial value will reduce the maximum emf by another factor of . Therefore, the maximum emf will be reduced by a factor of . The voltmeter reads the RMS value of the emf, which is just proportional to the maximum emf value. Since the initial voltmeter reading is 80 V, the voltmeter will read 10 V under the new conditions.

Correct Response: A. (Objective 0012) This question requires the examinee to apply Lenz's law and Faraday's law to explain and predict induced currents from a given changing magnetic field. As the wire coil rotates, the magnetic field through the coil changes, and by Faraday's law an emf is created in the loop. The voltmeter reads this emf. Faraday's law states that the emf depends on the rate of change of flux through the surface of the coil. The flux is
phi equals integral bold upper b dot d upper bold a equals upper b upper a cosine of open parenthesis omega t close parenthesis equals upper b x squared cosine of open parenthesis omega t close parenthesis where **B** is the magnetic field, *A* equals *x* squared is the area of the loop, and theta equals omega t *i*s the angle between the vector normal to the surface of the loop and **B**. Using Faraday's law to get the emf, epsilon , gives epsilon equals negative start fraction numerator d phi denominator d t end fraction equals upper b x squared omega sine of open parenthesis omega t close parenthesis. Reducing a side of the square loop to one half x reduces the area to one fourth x squared, so the maximum flux and the maximum emf will be reduced to one fourth of their previous values from this factor. Slowing the rotational speed, omega, to half its initial value will reduce the maximum emf by another factor of one half. Therefore, the maximum emf will be reduced by a factor of one eighth. The voltmeter reads the R M S value of the emf, which is just proportional to the maximum emf value. Since the initial voltmeter reading is 80 V, the voltmeter will read 10 V under the new conditions.

10. The diagram below shows a bar magnet of unknown polarity and a wire coil attached to a resistor.

a wire coiled around a cylinder and attached to a resistor. The current flow is shown as traveling counterclockwise righthand side of the cylinder. The magnet is shown with one pole marked X facing the right hand side of the cylinder.

In which of the following cases will positive current have the direction shown?

*X* is a north pole and the magnet is stationary.
*X* is a south pole and the magnet is stationary.
*X* is a north pole moving toward the coil.
*X* is a south pole moving toward the coil.

- Answer
**Correct Response: C.** (Objective 0012) This question requires the examinee to apply Lenz's law and Faraday's law to explain and predict induced currents from a given changing magnetic field. By Faraday's law, since there is a current in the coil and no external emf, the current is produced by a changing magnetic field, so the bar magnet must be in motion. From the right-hand rule, curling the fingers of the right hand in the direction of the induced current shows that the north pole of the induced field in the coil points to the right, and the south pole points to the left. Lenz's law states that when a changing magnetic field produces an emf, the polarity of the emf is produced to oppose that change in magnetic flux. The induced emf produces the current in the direction shown. Therefore, the north pole in the coil is produced to oppose the change in flux. In this case, the north pole of the magnetic field in the coil is produced to repel the pole of the bar magnet. Since like poles repel, *X* must be a north pole, and since the magnet is in motion as described above, *X* must be a north pole moving toward the coil to induce the current in the direction shown.

Correct Response: C. (Objective 0012) This question requires the examinee to apply Lenz's law and Faraday's law to explain and predict induced currents from a given changing magnetic field. By Faraday's law, since there is a current in the coil and no external emf, the current is produced by a changing magnetic field, so the bar magnet must be in motion. From the right-hand rule, curling the fingers of the right hand in the direction of the induced current shows that the north pole of the induced field in the coil points to the right, and the south pole points to the left. Lenz's law states that when a changing magnetic field produces an emf, the polarity of the emf is produced to oppose that change in magnetic flux. The induced emf produces the current in the direction shown. Therefore, the north pole in the coil is produced to oppose the change in flux. In this case, the north pole of the magnetic field in the coil is produced to repel the pole of the bar magnet. Since like poles repel, X must be a north pole, and since the magnet is in motion as described above, X must be a north pole moving toward the coil to induce the current in the direction shown.

11. Which of the following situations will produce an electromagnetic wave?

- an electric dipole
- an oscillating charge
- a constant current in a wire
- a nonuniform charge distribution

- Answer
**Correct Response: B.** (Objective 0012) This question requires the examinee to explain the electromagnetic nature of light. According to Maxwell's equations, electromagnetic waves are oscillating electric and magnetic fields. A changing electric field produces a changing magnetic field which produces a changing electric field and so on. Once electromagnetic waves are produced, they are perpetuated by the regenerative relationship between changing electric fields and changing magnetic fields. These changing fields propagate through a vacuum at the speed of light. An analysis of Maxwell's equations shows that these waves are generated when charges accelerate. Of the responses given, only the oscillating charge is undergoing acceleration.

Correct Response: B. (Objective 0012) This question requires the examinee to explain the electromagnetic nature of light. According to Maxwell's equations, electromagnetic waves are oscillating electric and magnetic fields. A changing electric field produces a changing magnetic field which produces a changing electric field and so on. Once electromagnetic waves are produced, they are perpetuated by the regenerative relationship between changing electric fields and changing magnetic fields. These changing fields propagate through a vacuum at the speed of light. An analysis of Maxwell's equations shows that these waves are generated when charges accelerate. Of the responses given, only the oscillating charge is undergoing acceleration.

12. A person would like to use a thin converging lens to generate a real image of an object placed 20 cm in front of the lens. The height of the object is 3.0 cm, and desired height of the image is 6.0 cm. What focal length should the lens have?

- 6.7 cm
- 9.0 cm
- 13.3 cm
- 40 cm

13. The diagram below shows a ray of light incident on a piece of glass. The index of refraction of air is *n* = 1.0, and that of the glass is *n* = 1.5. The angle between the incident and the reflected light ray is θ = 60°.

ray of light hitting a piece of glass. bottom portion is the glass and the top portion is air. the light hits the glass at a sixty degree angle and reflects off making a sixty degree angle labeled theta. There is a vertical dashed line intersecting the angle theta. below the glass the light ray continues in a negative fashion at a smaller angle phi off of the vertical line intersecting the larger angle at the top of the glass

Which of the following is closest to the measure of φ ?

- 15°
- 17°
- 19°
- 35°

- Answer
**Correct Response: C.** (Objective 0013) This question requires the examinee to describe the reflection, refraction, and transmission of light using Snell's law when light encounters a given object or a boundary between mediums having different indices of refraction. When light travels from one medium to another, it changes speed, which causes the wave to bend. Using the ray model of light, the amount of bending, or refraction, is given by Snell's law, *n*_{1} sin θ_{i} = *n*_{2} sin θ_{r}, where *n*_{1} is the index of refraction of the optical medium of the incident light, *n*_{2} is the index of refraction of the refracted light, and θ_{i} and θ_{r} are the angles of incidence and refraction, respectively. The angle of incidence is the angle made with the line normal (perpendicular) to the surface in the initial medium and the angle of refraction is with respect to the normal in the second medium. In the diagram above, the normal is the vertical line. Therefore, the angle of incidence is = 30° and the angle of refraction is φ . Using the values given and Snell's law, 1.0 sin 30° = 1.5 sin φ °, so sin φ = and sin−1 φ = , or θ = 19.47 or 19°.

Correct Response: C. (Objective 0013) This question requires the examinee to describe the reflection, refraction, and transmission of light using Snell's law when light encounters a given object or a boundary between mediums having different indices of refraction. When light travels from one medium to another, it changes speed, which causes the wave to bend.

Using the ray model of light, the amount of bending, or refraction, is given by Snell's law, *n* sub one sine theta sub i equals *n* sub two baseline sine theta sub r, where *n* sub one is the index of refraction of the optical medium of the incident light, *n* sub two is the index of refraction of the refracted light, and theta sub i* *and theta sub r* *are the angles of incidence and refraction, respectively. The angle of incidence is the angle made with the line normal (perpendicular) to the surface in the initial medium and the angle of refraction is with respect to the normal in the second medium.

In the diagram above, the normal is the vertical line. Therefore, the angle of incidence is theta over two equals 30 degrees and the angle of refraction is phi. Using the values given and Snell's law, 1.0 sine 30 degrees equals 1.5 sine phi degrees, so sine phi equals and sin minus 1 phi equals zero point five over one point five, or theta equals 19.47 or 19 degrees.

14. The ray diagram below shows the optics of an object on the optical axis of a concave mirror.

This ray diagram shows a horizontal axis starting on the left and ending in the center of a concave mirror. Four points labeled I, II, III, and IV are along the axis with I furthest from the mirror. At point I is an arrow normal to the axis labeled "object". A dashed ray is shown from the top of the arrow through point II at the axis and then reflecting from the mirror and back to the arrow. A second ray is shown from the arrow parallel to the axis to the mirror and reflected down through the axis at point IV. Point III is midway between points II and IV, at the same position on the axis where the two rays intersect below the axis.

Which of the following statements about this system is true?

- The mirror has a focal point at II and forms a virtual inverted image between I and II.
- The mirror has a focal point at II and forms a real upright image at IV.
- The mirror has a focal point at IV and forms a real inverted image at III.
- The mirror has a focal point between II and III and forms a virtual upright image at IV.

- Answer
**Correct Response: C.** (Objective 0013) This question requires the examinee to describe the location, types, and magnification of images formed by mirrors and lens systems using words, ray diagrams, and mathematical expressions. The focal point of a mirror is the point at which light rays parallel to the optical axis intersect the axis after being reflected from the mirror. According to the diagram, the focal point of the mirror is at point IV and light rays from the top of the object intersect beneath point III, forming an inverted image. This can be seen further by tracing a ray from the top of the object through point IV. The reflected ray will be parallel to the optical axis and intersect the other rays from the top of the image.

Correct Response: C. (Objective 0013) This question requires the examinee to describe the location, types, and magnification of images formed by mirrors and lens systems using words, ray diagrams, and mathematical expressions. The focal point of a mirror is the point at which light rays parallel to the optical axis intersect the axis after being reflected from the mirror. According to the diagram, the focal point of the mirror is at point I V and light rays from the top of the object intersect beneath point III, forming an inverted image. This can be seen further by tracing a ray from the top of the object through point IV. The reflected ray will be parallel to the optical axis and intersect the other rays from the top of the image.

15. An optical system consists of two converging lenses. The objective lens has a focal length of 1.0 cm and the eyepiece has a focal length of 4.0 cm. An object is located at a distance of 1.5 cm from the objective lens. The eyepiece is used as a magnifying glass.

In this ray diagram, there is an objective lens to the left parallel to an eyepiece lens to the right. They are separated by a distance d and shown with an axis drawn through the center of the two. An upward object arrow is shown to the left of the objective with one ray shown from the tip of the object arrow passing through the axis at the center of the objective lens, and connecting to the tip of a downward inverted image. A second ray is shown from the tip or the object arrow parallel to the axis to the objective lens then turning to meet the tip of the inverted image below the axis.

Between which of the following distances from the objective lens can the eyepiece be placed so that a person looking through the eyepiece will see a virtual inverted image of the object?

- 0.0 cm and 1.0 cm
- 1.0 cm and 1.5 cm
- 3.0 cm and 7.0 cm
- 7.0 cm and 16 cm

- Answer
**Correct Response: C.** (Objective 0013) This question requires the examinee to design a system of lenses and mirrors to produce a real or virtual image of a given magnification. In a system consisting of two lenses, the objective lens is used to form an image that is then used as the object of the second lens—in this case, the eyepiece. Since the object is at a distance greater than the focal length of the objective lens, the objective lens forms an inverted image of the object. The eyepiece is then used as a magnifying glass to create a virtual image of the inverted image. In order for a converging lens (the eyepiece) to create a virtual image of an object, the object must be located between the eyepiece and the focal point of the lens. Therefore, the eyepiece must be located between 0.0 cm and 4.0 cm from the inverted image formed by the objective lens. To find where the inverted image is located, apply the thin lens equation to the objective lens, , where f is the focal length and *s*_{o} and *s*_{i} are the distances from the objective lens of the object and the image, respectively. Since *f* = 1.0 cm and *s*_{o} = 1.5 cm, the equation becomes . Solving for *s*_{i} gives *s*_{i} = 3.0 cm, so the inverted image is formed
3.0 cm to the right of the objective lens. Since the eyepiece must be located between 0.0 cm and 4.0 cm from the inverted image to create a virtual image, the eyepiece must be located between 3.0 cm and 7.0 cm from the objective lens.

Correct Response: C. (Objective 0013) (Objective 0013) This question requires the examinee to design a system of lenses and mirrors to produce a real or virtual image of a given magnification. In a system consisting of two lenses, the objective lens is used to form an image that is then used as the object of the second lens—in this case, the eyepiece. Since the object is at a distance greater than the focal length of the objective lens, the objective lens forms an inverted image of the object. The eyepiece is then used as a magnifying glass to create a virtual image of the inverted image. In order for a converging lens (the eyepiece) to create a virtual image of an object, the object must be located between the eyepiece and the focal point of the lens. Therefore, the eyepiece must be located between 0.0 c m and 4.0 c m from the inverted image formed by the objective lens. To find where the inverted image is located, apply the thin lens equation to the objective lens, one over f equals one over s sub zero baseline plus one over s sub i, where f is the focal length and *s* sub zero and *s *sub i are the distances from the objective lens of the object and the image, respectively. Since *f* equals 1.0 c m and *s *sub zero equals 1.5 c m, the equation becomes one over one point zero equals one over one point five plus one over s sub i. Solving for *s *sub i gives *s* sub i equals 3.0 c m, so the inverted image is formed
3.0 cm to the right of the objective lens. Since the eyepiece must be located between 0.0 c m and 4.0 c m from the inverted image to create a virtual image, the eyepiece must be located between 3.0 c m and 7.0 c m from the objective lens.

16. Unpolarized light of intensity *I* is incident on a system of two polarizers. Two photometers are positioned as shown in the diagram below. The axis of the second polarizer is aligned at 60° with respect to the vertical axis of the first polarizer.

This diagram shows incident light rays passing through a series of two polarizers. The second polarization in the second is rotated sixty degrees from the first. Photometer one is between the two polarizers, and Photometer two is reading after the light passes through both the polarizers.

What intensity will the two photometers register?

- Photometer 1 will read
*I* and photometer 2 will read .
- Photometer 1 will read
*I* and photometer 2 will read .
- Photometer 1 will read and photometer 2 will read .
- Photometer 1 will read and photometer 2 will read .

- Answer
**Correct Response: D.** (Objective 0014) This question requires the examinee to demonstrate knowledge of measurements and calculations used to determine the intensity and polarization of a given light source. Electromagnetic waves consist of oscillating electric and magnetic fields that propagate through space at the speed of light, *c*. The waves are transverse: the fields oscillate in a direction perpendicular to the direction of wave propagation. Unpolarized light consists of a superposition of many light waves that have their electric and magnetic field vectors pointing in random directions perpendicular to the wave propagation. Since the magnitude of the magnetic field is small compared to the electric field, it is the electric field that interacts with matter most strongly. Polarizing filters are typically made of plastic that contains tiny crystals arranged in the same direction so that they transmit only the component of the electric field parallel to the direction of alignment; the perpendicular component of the electric field is absorbed. Each electric field vector can be resolved to a component that is parallel and a component that is perpendicular to the polarizing axis. The parallel component will pass through the filter and the perpendicular component will be absorbed. Since the electric field vectors are randomly aligned in unpolarized light, half of the light intensity corresponds to the electric field components parallel to the polarizing axis. Therefore, the photometer between the filters will read . Malus's law can be used to find the amount of light transmitted through the second filter. The intensity of transmitted light is *I*_{t} = *I*_{0} cos^{2} θ , where *I*_{0} is the intensity of the polarized beam between the two filters and θ is the angle between the two polarizing axes. Since θ = 60° and cos 60° = , cos^{2} θ = , and
*I*_{0} = , the second photometer will register an intensity of .

Correct Response: D. (Objective 0014) This question requires the examinee to demonstrate knowledge of measurements and calculations used to determine the intensity and polarization of a given light source. Electromagnetic waves consist of oscillating electric and magnetic fields that propagate through space at the speed of light, *c*. The waves are transverse: the fields oscillate in a direction perpendicular to the direction of wave propagation. Unpolarized light consists of a superposition of many light waves that have their electric and magnetic field vectors pointing in random directions perpendicular to the wave propagation. Since the magnitude of the magnetic field is small compared to the electric field, it is the electric field that interacts with matter most strongly. Polarizing filters are typically made of plastic that contains tiny crystals arranged in the same direction so that they transmit only the component of the electric field parallel to the direction of alignment; the perpendicular component of the electric field is absorbed. Each electric field vector can be resolved to a component that is parallel and a component that is perpendicular to the polarizing axis. The parallel component will pass through the filter and the perpendicular component will be absorbed. Since the electric field vectors are randomly aligned in unpolarized light, half of the light intensity corresponds to the electric field components parallel to the polarizing axis. Therefore, the photometer between the filters will read one half i . Malus's law can be used to find the amount of light transmitted through the second filter. The intensity of transmitted light is *I *sub t equals *I* sub zero cosine squared theta , where *I* sub zero is the intensity of the polarized beam between the two filters and theta is the angle between the two polarizing axes. Since theta equals 60 degrees; and cosine 60 degrees; equals one half , cosine squared theta equals one fourth, and *I* sub zero equals one hald i, the second photometer will register an intensity of one half i times one fourth equals one eighth i.

17. The diagram below shows two monochromatic, coherent light sources that are separated by a small distance *d*. Light from the two sources strikes the screen at point *A*. The light sources are located at distances of *s*_{1} and *s*_{2} from point *A*.

two light sources, S sub one and S sub two, emitted from two slits a distance d apart. The two light sources converge on a screen parallel to the sources at one point A.

Which of the following best explains the necessary condition for point *A* on the screen to be a region of maximum light intensity?

- when the ratio of
*d* to *s*_{1} equals λ
- when d is of the same order of magnitude as λ
- when the sum of
*d* and *s*_{1} is equal to *s*_{2}
- when the difference between
*s*_{2} and *s*_{1} is an integral multiple of λ

- Answer
**Correct Response: D.** (Objective 0014) This question requires the examinee to describe the interaction of monochromatic light with a given single slit, a pair of parallel slits, and with thin films. The light waves are monochromatic, so they all have the same wavelength. The light is also coherent, which means that the light from each source is in phase. According to the superposition principle, two light waves that travel along paths *s*_{1} and *s*_{2} will add when they arrive at point A. If the waves add constructively, there will be a bright band on the screen. If the waves add destructively, there will be a dark band (no light) on the screen. Since the waves are in phase, they will add constructively if the wave amplitudes are aligned. This will occur when the difference in the length of the paths taken (*s*_{2}–*s*_{1}) is an integral multiple of the wavelength. The waves will add destructively if the path lengths differ by one-half of a wavelength.

Correct Response: D. (Objective 0014) This question requires the examinee to describe the interaction of monochromatic light with a given single slit, a pair of parallel slits, and with thin films. The light waves are monochromatic, so they all have the same wavelength. The light is also coherent, which means that the light from each source is in phase. According to the superposition principle, two light waves that travel along paths *s* sub one and *s* sub twowill add when they arrive at point A. If the waves add constructively, there will be a bright band on the screen. If the waves add destructively, there will be a dark band (no light) on the screen. Since the waves are in phase, they will add constructively if the wave amplitudes are aligned. This will occur when the difference in the length of the paths taken open parenthesis *s* sub two minus *s* sub one close parenthesisis an integral multiple of the wavelength. The waves will add destructively if the path lengths differ by one-half of a wavelength.

18. Light from a laser passes through a single slit of width *a*, creating the light pattern shown below.

five adjacent flattened ovals. The largest is in the center, and is labeled with a width of delta y. Two equivalent midsized ovals are on either side, and outer most are two equivalent small ovals.

For which color of laser light will Δ*y* be greatest?

- red
- yellow
- green
- blue

- Answer
**Correct Response: A.** (Objective 0014) This question requires the examinee to apply the principle of superposition to a system of slits to explain and predict how changing the slit width, the slit separation, and the wavelength of the incident light affects the resulting light pattern. For a single slit, the regions in which the light intensity is minimum are given by *d* sin θ = *m*λ , where *d* is the width of the slit, θ is the angular displacement of the beam, *m* is the order of the minimum, and λ is the wavelength. The distance Δ*y* corresponds to the distance between the first two minimums when *m* = 1. From this it follows that Δ*y* will increase as the wavelength of the light increases—the longer the wavelength of light, the greater the distance between the first two minimums. Of the choices given, red light has the longest wavelength.

Correct Response: A. (Objective 0014) This question requires the examinee to apply the principle of superposition to a system of slits to explain and predict how changing the slit width, the slit separation, and the wavelength of the incident light affects the resulting light pattern. For a single slit, the regions in which the light intensity is minimum are given by *d* sine theta equals *m* times lambda, where *d* is the width of the slit, theta is the angular displacement of the beam, *m* is the order of the minimum, and lambda is the wavelength. The distance delta times *y* corresponds to the distance between the first two minimums when *m* equals 1. From this it follows that delta times *y* will increase as the wavelength of the light increases—the longer the wavelength of light, the greater the distance between the first two minimums. Of the choices given, red light has the longest wavelength.

19. An ideal gas has a heat capacity of 20 J/(mol•K). One mole of the gas is in a cylinder and absorbs 1000 J of heat and lifts a 10 kg mass a vertical distance of 2 m while expanding. If the initial temperature of the gas is 300 K, what is the approximate final temperature of the gas?

An ideal gas has a heat capacity of twenty joules per mole kelvin. One mole of the gas is in a cylinder and absorbs one thousand joules of heat and lifts a ten kilogram mass a vertical distance of two meters while expanding. If the initial temperature of the gas is three hundred kelvins, what is the approximate final temperature of the gas?

A. two hundred and fifty-one kelvins

B. two hundred and sixty kelvins

C. three hundred forty kelvins

D. three hundred forty nine kelvins

- 251 K
- 260 K
- 340 K
- 349 K

20. A heat engine consists of 1.0 mol of an ideal gas operating in a Carnot cycle. In state I in the graph below, the gas has a pressure of 20,000 Pa and a volume of 0.50 m^{3}. In state III the gas has a pressure of 5,000 Pa and a volume of 1.0 m^{3}.

graph with x axis Volume and y axis Pressure. Four points labelled I, II, III, and IV are connected in a non intersecting, continuous cycle of roughly parallelogram shape. Point I has lowest V and highest P. Point II has a higher V and lower P. Point III has the highest V and the lowest P. Point IV has a lower V and higher P.

What is the efficiency of the heat engine?

- 20%
- 30%
- 50%
- 70%

- Answer
**Correct Response: C.** (Objective 0015) This question requires the examinee to use the first law of thermodynamics to explain and predict the thermal efficiency and the changes in pressure, temperature, and volume of a monatomic ideal gas operating in a Carnot cycle. The efficiency of a heat engine operating between two heat reservoirs, *T*_{h} and *T*_{c}, is given by Since the processes from I to II and from III to IV are isothermals, the temperature of the gas in state I and state III is the temperature of the heat reservoirs and can be found using the ideal gas law, *PV* = *nRT*, where *R* = 8.31 J/mol-K. It follows that = 1,203 K, = 601.7 K, and the efficiency is *e* = 50%.

Correct Response: C. (Objective 0015) This question requires the examinee to use the first law of thermodynamics to explain and predict the thermal efficiency and the changes in pressure, temperature, and volume of a monatomic ideal gas operating in a Carnot cycle. The efficiency of a heat engine operating between two heat reservoirs, T sub h and T sub c, is given by e equals start fraction numerator t sub h baseline minus t sub c denominator t sub h end fraction. Since the processes from I to II and from III to IV are isothermals, the temperature of the gas in state I and state III is the temperature of the heat reservoirs and can be found using the ideal gas law, P V equals n R T, where R equals 8.31 J slash mol K. It follows that t sub h equals start fraction numerator p sub h baseline v sub h denominator n r end fraction equals start fraction numerator open parenthesis twenty thousand k upper p a close parenthesis open parenthesis zero point five zero m cubed close parenthesis denominator open parenthesis one point zero mol close parenthesis open parenthesis eight point three one j slash mol k close parenthesis end fraction equals 1,203 K, t sub c equals start fraction numerator p sub c baseline v sub c denominator n r end fraction equals start fraction numerator open parenthesis five thousand k upper p a close parenthesis open parenthesis one point zero m cubed close parenthesis denominator open parenthesis one point zero mol close parenthesis open parenthesis eight point three one j slash mol k close parenthesis end fraction equals 601.7 K, and the efficiency is e equals fifty percent.

21. If the pressure of an ideal gas is doubled while the volume of the gas remains constant, then the average speed of the molecules will increase by a factor of:

- √2.
- 2.

- Answer
**Correct Response: C.** (Objective 0015) This question requires the examinee to use the kinetic-molecular model of matter to explain common physical changes (e.g., changes in temperature or pressure of a gas, phase changes). The average kinetic energy of the molecules in an ideal gas is proportional to the temperature of the gas. If the pressure of the gas is doubled while the volume remains constant, the temperature can be found using the idea gas law, PV = nRT. Since n, R, and V are constant, then = constant and , , and T2 = 2T1; that is, the temperature doubles. Since temperature is proportional to the average kinetic energy of the molecules, , , and *v*_{2} = √2*v*_{1}. Thus, the average speed of the molecules increases by a factor of √2.

Correct Response: C. (Objective 0015) This question requires the examinee to use the kinetic-molecular model of matter to explain common physical changes (e.g., changes in temperature or pressure of a gas, phase changes). The average kinetic energy of the molecules in an ideal gas is proportional to the temperature of the gas. If the pressure of the gas is doubled while the volume remains constant, the temperature can be found using the idea gas law, P V equals n R T. Since n, R, and V are constant, then p over t equals constant and p sub one over t sub one equals p sub two over t sub two, p sub one over t sub one equals p sub two over t sub two, p sub one over t sub one equals two p sub one over t sub two, and T 2 equals 2 T 1; that is, the temperature doubles. Since temperature is proportional to the average kinetic energy of the molecules, one half m v sub one baseline squared equals one half m v sub two baseline squared, v sub two baseline squared equals two v sub one baseline squared, and *v* sub two equals square root of 2 *v* sub one Thus, the average speed of the molecules increases by a factor of square root of 2.

22. A 10 Ω resistor is placed in an insulated calorimeter containing 0.25 kg of water at 20°C. The resistor is connected to a constant current source, and after 64 s the temperature of the water is 30°C. No heat is lost to the surroundings and the heat capacity of the calorimeter and resistor can be ignored. What is the current in the resistor?

- 4 A
- 8 A
- 16 A
- 32 A

- Answer
**Correct Response: A.** (Objective 0015) This question requires the examinee to use the first law of thermodynamics to explain and predict the final temperature of a given thermally isolated system and the transfer of heat into or out of a given system. As current passes through the resistor, heat is dissipated due to Joule heating. The power dissipated is *P* = *IV*, which can be combined with Ohm's law, *V* = *IR*, to get *P* = *I*^{2}*R*. Since power is the rate of change of energy, , multiplying the power by the time interval gives an expression for the energy dissipated, Δ*E* = Δ*tI*^{2}*R*. By the first law of thermodynamics, this must equal the heat gained by the water, *Q* = *cm*Δ*T*. Setting these two expressions equal using *c* = 4.19 × 103 J/(kg-°C), substituting values, and solving for I results in *I* = 4 A.

Correct Response: A. (Objective 0015) This question requires the examinee to use the first law of thermodynamics to explain and predict the final temperature of a given thermally isolated system and the transfer of heat into or out of a given system. As current passes through the resistor, heat is dissipated due to Joule heating. The power dissipated is *P *equals *I V*, which can be combined with Ohm's law, *V* equals *I R*, to get *P* equals *I* squared *R*. Since power is the rate of change of energy, p equals delta e over delta t, multiplying the power by the time interval gives an expression for the energy dissipated, delta *E* equals delta *t I* squared *R*. By the first law of thermodynamics, this must equal the heat gained by the water, *Q* equals *c m* delta *T*. Setting these two expressions equal using *c* equals 4.19 times 103 J slash open parenthesis k g - degrees celsius close parenthesis, substituting values, and solving for I results in *I* equals 4 A.

23. A Geiger counter placed near a radioactive source registers 80 counts per minute when adjusted for background radiation. When a piece of paper is placed between the source and the Geiger counter window, the number drops to 0 counts per minute. This indicates that the source is emitting:

- alpha particles only.
- beta particles only.
- gamma particles only.
- beta and gamma particles only.

- Answer
**Correct Response: A.** (Objective 0016) This question requires the examinee to demonstrate knowledge of measurements used to detect nuclear radiation in the environment. A Geiger counter measures ionizing radiation. Most radioactive sources emit alpha, beta, or gamma radiation. Of these types of radiation, alpha particles have the lowest energy and consist of two protons and two neutrons. These particles are generally stopped by a piece of paper. Beta particles are electrons emitted by the nucleus and have greater energy. These particles are usually stopped by a sheet of aluminum foil. Gamma particles are high-energy photons that can only be stopped by a thick sheet of lead or a large slab of concrete. Since the number registered by the Geiger counter dropped to 0 counts per minute when a piece of paper was placed between the window of the counter and the source, the source is emitting only alpha particles.

Correct Response: A. (Objective 0016) This question requires the examinee to demonstrate knowledge of measurements used to detect nuclear radiation in the environment. A Geiger counter measures ionizing radiation. Most radioactive sources emit alpha, beta, or gamma radiation. Of these types of radiation, alpha particles have the lowest energy and consist of two protons and two neutrons. These particles are generally stopped by a piece of paper. Beta particles are electrons emitted by the nucleus and have greater energy. These particles are usually stopped by a sheet of aluminum foil. Gamma particles are high-energy photons that can only be stopped by a thick sheet of lead or a large slab of concrete. Since the number registered by the Geiger counter dropped to 0 counts per minute when a piece of paper was placed between the window of the counter and the source, the source is emitting only alpha particles.

24. The reaction below shows the radioactive decay of potassium-40 into calcium-40 and an unidentified particle, *X*.

Upper k atomic number nineteen and mass of forty produces upper c a with atomic number of twenty and a mass of forty plus upper x

What is the identity of *X* in this reaction?

- a positron
- a beta particle
- a proton
- an alpha particle

- Answer
**Correct Response: B.** (Objective 0016) This question requires the examinee to use conservation principles to explain observed changes in matter in a given nuclear process. According to the reaction, the number of protons in the nucleus of the potassium atom increases by one while the atomic mass remains the same. Therefore, a neutron must have decayed into a proton and an electron. This is known as beta decay, and beta particles are electrons that arise from the decay of a neutron into a proton and an electron. Emission of a beta particle from a radioisotope will increase the number of protons in the nucleus by one and decrease the number of neutrons by one, maintaining the same number of nucleons.

Correct Response: B. (Objective 0016) This question requires the examinee to use conservation principles to explain observed changes in matter in a given nuclear process. According to the reaction, the number of protons in the nucleus of the potassium atom increases by one while the atomic mass remains the same. Therefore, a neutron must have decayed into a proton and an electron. This is known as beta decay, and beta particles open parenthesis beta with a mass of zero and an atomic number of negative one close parenthesis are electrons that arise from the decay of a neutron into a proton and an electron. Emission of a beta particle from a radioisotope will increase the number of protons in the nucleus by one and decrease the number of neutrons by one, maintaining the same number of nucleons.

25. According to the standard model, which of the following best describes the mechanism of the strong, weak, and electromagnetic interactions?

- The interactions generate energy through the decay of dark matter.
- The interactions result from the exchange of bosons between particles.
- The interactions are a manifestation of gravity in different reference frames.
- The interactions are produced by the pairing of quarks and antiquarks.

- Answer
**Correct Response: B.** (Objective 0016) This question requires the examinee to use the standard model to explain qualitatively the observed interactions between atomic or subatomic particles in simple situations. The standard model describes the strong, weak, and electromagnetic interactions as forces that are mediated by the exchange of particles called bosons. The photon is an example of a boson, and it mediates the electromagnetic force.

Correct Response: B. (Objective 0016) This question requires the examinee to use the standard model to explain qualitatively the observed interactions between atomic or subatomic particles in simple situations. The standard model describes the strong, weak, and electromagnetic interactions as forces that are mediated by the exchange of particles called bosons. The photon is an example of a boson, and it mediates the electromagnetic force.

26. A GPS (global positioning system) satellite has an orbital speed of about 14,000 kg/hr, or 3,889 m/s. According to the theory of special relativity, compared to the stationary clock on the earth, the clock in the satellite will measure a second as being:

- longer and therefore run slow.
- longer and therefore run fast.
- shorter and therefore run slow.
- shorter and therefore run fast.

- Answer
**Correct Response: A.** (Objective 0016) This question requires the examinee to use words, diagrams, and mathematical relationships to describe time dilation, length contraction, and momentum and energy of an object at a given velocity. According to the theory of special relativity, a clock in an inertial reference frame moving at a constant speed with respect to a clock in a stationary reference frame will undergo time dilation. This means that a second measured in the moving clock will be slightly longer than a second measured in the stationary clock. This will result in the clock in the satellite running more slowly than the clock in the stationary frame.

Correct Response: A. (Objective 0016) This question requires the examinee to use words, diagrams, and mathematical relationships to describe time dilation, length contraction, and momentum and energy of an object at a given velocity. According to the theory of special relativity, a clock in an inertial reference frame moving at a constant speed with respect to a clock in a stationary reference frame will undergo time dilation. This means that a second measured in the moving clock will be slightly longer than a second measured in the stationary clock. This will result in the clock in the satellite running more slowly than the clock in the stationary frame.

27. Which of the following best characterizes a basic property of semiconductors?

- Electrons in a semiconductor are free to scatter from atom to atom, resulting in a low value of conductivity.
- Semiconductors contain a region where the valence band energies overlap with the conduction band.
- Electrons in the atoms are paired with electron holes that pass through the material with almost zero resistance.
- Electrons with sufficient energy can transition a narrow energy gap between the valence band and the conduction band.

28. An arrangement of electric and magnetic fields is used as a velocity selector to determine the speed, *v*, of an electron of mass *m* moving in one dimension. The speed of the electron can be determined with an accuracy of ±10% of *v*. Which of the following gives the uncertainty in the position of the electron?

- Answer
**Correct Response: B.** (Objective 0017) This question requires the examinee to use the Heisenberg uncertainty principle to predict the lower limit of size, momentum, energy, or time expected in a given atomic or subatomic process. The uncertainty principle expresses a relationship between measuring the momentum and position of a particle simultaneously. One form of the relationship states that Δ*x*Δ*p* ≥ *h*. Since the accuracy in the speed, *v*, of the electron is 10%, then the uncertainty in the electron's speed is Δ*v* = (*v* + 0.1*v*) − (*v* − 0.1*v*) = 0.2*v*. Since p = m*v*, Δ*p* = *m*Δ*v*. Substituting this into the uncertainly principle and solving for Δx gives .

Correct Response: B. (Objective 0017) This question requires the examinee to use the Heisenberg uncertainty principle to predict the lower limit of size, momentum, energy, or time expected in a given atomic or subatomic process. The uncertainty principle expresses a relationship between measuring the momentum and position of a particle simultaneously. One form of the relationship states that delta *x* delta *p* greater then or equal to *h*. Since the accuracy in the speed, *v*, of the electron is ten percent, then the uncertainty in the electron's speed is delta *v* equals open parenthesis* v * plus 0.1 *v* close parenthesis minus open parenthesis *v* minus 0.1 *v* close parenthesis equals 0.2 *v*. Since p equals m *v*, delta *p* equals *m* delta *v*. Substituting this into the uncertainly principle and solving for delta x gives delta x greater than or equals to start fraction numerator h denominator zero point two m v end fraction.

29. The electron energy level diagram for an atom consisting of a single electron that is only allowed to occupy three states is shown below.

electron energy level diagram. Vertical line with three arms coming out to the left. starting at the bottom we have n equals one then a large distance up to n equals two then a short distance up to n equals three

Which of the following must be known in order to calculate the frequency of the lines in the emission spectrum of this atom?

- the spin of the electron in each state
- the energy difference between the states
- the ground state energy of the electron
- the uncertainty in momentum of the electron in each state

- Answer
**Correct Response: B.** (Objective 0017) This question requires the examinee to demonstrate knowledge of calculations used to analyze the emission spectrum of a given gas. The atomic spectrum of an isotope is generated by electrons in an atom undergoing transitions from one energy state to another. For example, the frequency of light emitted as the electron transitions from the *n* = 3 excited state to the *n* = 2 excited state can be calculated using the equation *E*_{3} − *E*_{2} = *hf*. Since the value of Planck's constant (*h*) is known, the frequency, *f*, can be determined for each of the three possible electron transitions if the energy difference between the states is known.

Correct Response: B. (Objective 0017) This question requires the examinee to demonstrate knowledge of calculations used to analyze the emission spectrum of a given gas. The atomic spectrum of an isotope is generated by electrons in an atom undergoing transitions from one energy state to another. For example, the frequency of light emitted as the electron transitions from the n equals 3 excited state to the n equals 2 excited state can be calculated using the equation E sub three minus E sub two equals h f. Since the value of Planck's constant (h) is known, the frequency, f, can be determined for each of the three possible electron transitions if the energy difference between the states is known.

30. A scientist performs an experiment with a microscope that uses photons to determine the position of an electron. The electron is in a potential well of width *L*, where L is
of the order of magnitude of the diameter of an atom. The scientist starts with electromagnetic radiation of wavelength λ = *L* and progressively uses radiation of smaller and smaller wavelengths. Which of the following best describes what will occur as the size of the wavelength decreases?

- The energy of the incident photons will decrease.
- The de Broglie wavelength of the electron will increase.
- The electron will drop to lower energy levels in the well.
- The momentum of the photons will disturb the electron's position.

- Answer
**Correct Response: D.** (Objective 0017) This question requires the examinee to use the quantum nature of light and matter and the conservation of energy and momentum to explain qualitatively the observed interaction between photons and matter in a given situation. Photons have momentum, and when they interact with an electron through a collision, some of the momentum will be transferred to the electron, further disturbing its position. The momentum of a photon is given by As the wavelength of the photons decreases, the momentum of the photons increases. These photons will scatter off the electron and disturb its position.

Correct Response: D. (Objective 0017) This question requires the examinee to use the quantum nature of light and matter and the conservation of energy and momentum to explain qualitatively the observed interaction between photons and matter in a given situation. Photons have momentum, and when they interact with an electron through a collision, some of the momentum will be transferred to the electron, further disturbing its position. The momentum of a photon is given by p equals h over lambda. As the wavelength of the photons decreases, the momentum of the photons increases. These photons will scatter off the electron and disturb its position.

## Answer Key