Modern Physics — Chapter Practice, Grade 12

Compare wave–particle evidence, relativistic limits, and nuclear mass-energy accounting.

Problem 1

Photon 5 eV, work function 2 eV. Find Kmax.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

3 eV

5−2=3.

Problem 2

Why does brighter below-threshold light not eject electrons in the basic model?
Answer and explanation

Each photon lacks the required energy; adding more such photons does not change the single-photon energy.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

Using hc=1240 eV·nm, find photon energy at 620 nm.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

2 eV

1240/620=2.

Problem 4

Which observation supports a wave model and which supports a photon model?
Answer and explanation

Interference supports wave superposition; threshold-dependent photoemission supports quantized energy exchange.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 5

With work function 2.4 eV and photons of 2.0 eV, is emission predicted by the single-photon model? Show the reasoning, not just the conclusion.
Hints

Below threshold, increased intensity alone does not supply more energy to each photon.

Answer and explanation

No. Each photon is below threshold; do not report negative kinetic energy as an emitted-electron result.

Problem 6

For fixed above-threshold frequency, what can increased light intensity change in the ideal photoelectric model?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

It can increase emission rate/current, while maximum electron kinetic energy is set by photon energy minus work function.

Problem 7

Maximum kinetic energy and stopping potential need an electron accountMaximum kinetic energy and stopping potential need an electron accountQuantitySupplied valueCalculationPhoton energy5.0 eVGiven frequency via hfWork function3.0 eVEscape thresholdKmax2.0 eV5.0−3.0Stopping-potential magnitude2.0 VeVs=KmaxMaximum energy is not every electron’s energy.
Original photoelectric energy/voltage account for the selected threshold model.
Interpret the supplied visual. What is 2.0 eV in joules?
Answer and explanation

3.204×10⁻¹⁹ J per electron, rounded appropriately for the supplied data.

Photoelectrons can emerge with a range of kinetic energies; the ideal relation specifies the maximum. A retarding stopping potential magnitude Vs is sufficient to stop even those fastest electrons when eVs=Kmax in the stated setup. Converting eV to joules uses 1 eV=1.602×10⁻¹⁹ J. Keep electron charge magnitude separate from the signed field/voltage convention when describing the apparatus, and do not confuse a 2 eV energy with a 2 V potential.

3.204×10⁻¹⁹ J per electron, rounded appropriately for the supplied data.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 8

A material has work function 2.2 eV. Photons carry 3.7 eV. Find Kmax in eV and joules, and stopping-potential magnitude.
Answer and explanation

Kmax=1.5 eV=2.403×10⁻¹⁹ J; stopping-potential magnitude=1.5 V. At fixed photon energy, increasing intensity changes available electron number rather than this maximum.

Kmax=1.5 eV=2.403×10⁻¹⁹ J; stopping-potential magnitude=1.5 V.

At fixed photon energy, increasing intensity changes available electron number rather than this maximum.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 9

Proper time 4 s andγ=2. Find dilated interval.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

8 s

2×4=8.

Problem 10

Why does Newtonian mechanics work well for walking-speed motion?
Answer and explanation

v/c is very small, so relativistic corrections are negligible for ordinary precision.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 11

A proper interval is 12 s and γ=1.25. Find other-frame interval.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

15 s

1.25×12=15.

Problem 12

Does a photon’s zero rest mass mean zero energy?
Answer and explanation

No. Its energy is hf and its momentum E/c.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 13

Find γ for v=0.80c. Show the reasoning, not just the conclusion.
Hints

Specify which events and clock define proper time.

Answer and explanation

1/√0.36=1/0.6≈1.67.

Problem 14

Why is classical mechanics often adequate for everyday speeds?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

When v is much smaller than c, relativistic corrections are extremely small within ordinary measurement precision.

Problem 15

Length contraction uses simultaneous endpoint positions in one frameLength contraction uses simultaneous endpoint positions in one frameRest length L₀=10 mAt0.60c: L=L₀/γ=8 mMoving-frame endpoints must be measured simultaneously.
Original rest/moving length schematic for parallel motion, not a literal visual photograph.
Interpret the supplied visual. Why is an ordinary photograph not automatically a direct length-contraction measurement?
Answer and explanation

Its image combines light emitted at different times and travel paths; simultaneous endpoint coordinates need separate reconstruction.

Proper length L₀ is measured in the object’s rest frame. An inertial frame where it moves parallel to its length measures L=L₀/γ, using endpoint positions at the same time in that measuring frame. The apparent photograph is a different problem because light from different parts can arrive after different travel times. Only the motion-parallel component contracts in this simple comparison; the object is not described as mechanically crushed by motion.

Its image combines light emitted at different times and travel paths; simultaneous endpoint coordinates need separate reconstruction.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 16

A process has proper duration 3.0 μs and moves at 0.80c relative to a lab. Find γ and lab duration; a rest 10 m object moving parallel at that speed has what lab length?
Answer and explanation

γ=1/√0.36=5/3. Lab duration=5.0 μs; length=10/(5/3)=6.0 m, with simultaneous endpoints in the lab.

γ=1/√0.36=5/3.

Lab duration=5.0 μs; length=10/(5/3)=6.0 m, with simultaneous endpoints in the lab.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 17

Mass difference.002u with 931.5 MeV/u gives what energy?
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

1.863 MeV

.002×931.5.

Problem 18

Why distinguish activity from dose?
Answer and explanation

Activity is source decay rate; dose depends on energy deposited per mass and exposure conditions.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 19

Using 931.5 MeV/u, find energy for .004 u modeled mass loss.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

3.726 MeV

.004×931.5=3.726.

Problem 20

Does rest-mass reduction violate energy conservation?
Answer and explanation

No; the difference appears as other forms of energy, with total energy conserved.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 21

Find energy for mass difference 1.0×10⁻³⁰ kg using the same c. Show the reasoning, not just the conclusion.
Hints

Compare complete initial and final rest-mass accounting.

Answer and explanation

9.0×10⁻¹⁴ J.

Problem 22

Why do nuclear reaction equations require both nucleon and charge-number checks?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

They constrain particle accounting; emitted particles must be included to balance the complete reaction model.

Problem 23

Energy per event, event rate and total output are different scalesEnergy per event, event rate and total output are different scalesScaleCalculationResultOne event2 MeV×1.602×10⁻¹³ J/MeV3.204×10⁻¹³ J10¹² eventsMultiply by event count0.3204 JPowerEnergy / elapsed timeNeeds a timeintervalLarge energy per event does not alone determine device output.
Original event-count scaling for a hypothetical 2 MeV channel, not an operating reactor record.
Interpret the supplied visual. What power corresponds to 0.3204 J released steadily in 2 s?
Answer and explanation

0.1602 W before any useful-output losses are applied.

A reaction’s energy per event must be multiplied by event number to obtain total energy. Power additionally needs event rate or duration and a useful-output efficiency. A large nuclear energy per event therefore does not by itself specify a power plant’s output or a sample’s radiation signal. Include how much material participates, what fraction reacts, and where released energy goes. Compare like energy services rather than isolated per-event numbers.

0.1602 W before any useful-output losses are applied.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 24

A supplied reaction releases 0.0015 u of rest-mass equivalent per event. Estimate energy in MeV, then explain why a detector count rate cannot directly give absorbed dose.
Answer and explanation

Energy≈0.0015×931.5=1.397 MeV/event. Dose also needs energy deposited, exposed mass, radiation characteristics and geometry; count rate includes detector efficiency/background and is not dose.

Energy≈0.0015×931.5=1.397 MeV/event.

Dose also needs energy deposited, exposed mass, radiation characteristics and geometry; count rate includes detector efficiency/background and is not dose.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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