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A Level H1 Physics Modern Physics Quiz

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Questions

A-Level Physics H1 Quiz - Modern Physics

Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 50

Duration: 45 minutes
Total Marks: 50

Instructions:

Answer ALL questions in the spaces provided.
Show all working for calculation questions.
Where appropriate, use g = 9.81 m s⁻², c = 3.00 × 10⁸ m s⁻¹, h = 6.63 × 10⁻³⁴ J s, e = 1.60 × 10⁻¹⁹ C.
Marks are indicated in brackets [ ].

Section A: Short Answer and Structured Response (Questions 1-5, 10 marks)

Answer all questions in this section.

1. State what is meant by the photoelectric effect. [2]

2. Define the term "work function" of a metal. [2]

3. Explain why the photoelectric effect cannot be explained by the wave model of light. [3]

4. State two observations from photoelectric effect experiments that support the photon model of light. [2]

5. A metal surface has a work function of 2.30 eV. Calculate the threshold frequency for this metal. [3]

Section B: Core Concepts and Equations (Questions 6-10, 10 marks)

Answer all questions in this section.

6. Explain what is meant by "stopping potential" in the context of the photoelectric effect. [2]

7. State Einstein's photoelectric equation and define each symbol used. [3]

8. Ultraviolet light of wavelength 200 nm is incident on a metal surface with work function 4.50 eV. Determine whether electrons will be emitted from the surface. Show your working. [3]

9. A clean metal surface is illuminated with light of wavelength 5.50 × 10⁻⁷ m. The stopping potential is measured to be 0.38 V. Calculate the work function of the metal, in electronvolts. [2]

10. A metal surface has a work function Φ. When light of frequency 2f₀ (where f₀ is the threshold frequency) is incident on the surface, electrons are emitted with maximum kinetic energy K. Determine the maximum kinetic energy of electrons emitted when light of frequency 3f₀ is used, in terms of K. [2]

Section C: Calculations and Data Analysis (Questions 11-15, 20 marks)

Answer all questions in this section.

11. Monochromatic light of frequency 8.00 × 10¹⁴ Hz is incident on a metal surface with work function 2.00 eV.

(a) Calculate the energy of a single photon of this light, in joules. [2]

(b) Calculate the maximum kinetic energy of the emitted electrons, in joules. [2]

12. In a photoelectric experiment, the maximum kinetic energy E_max of emitted electrons is measured for different wavelengths λ of incident light. The data obtained is shown below.

λ / nm	E_max / eV
450	0.55
400	0.90
350	1.35
300	1.95
250	2.75

(a) Calculate the frequency corresponding to each wavelength. Tabulate your results. [3]

(b) Plot a graph of E_max (y-axis) against frequency f (x-axis) on the grid provided. [4]

(d) Use your graph to determine the work function of the metal. [2]

13. In a photoelectric experiment, the stopping potential V_s is measured for different frequencies f of incident light. The results are plotted on a graph of V_s against f.

(a) State the physical quantity represented by the gradient of this graph. [1]

(b) State the physical quantity represented by the x-intercept of this graph. [1]

14. A student performs a photoelectric experiment using two different light sources. Source A has twice the intensity but the same frequency as Source B. Both frequencies are above the threshold frequency of the metal.

Compare and explain the photoelectric current and the maximum kinetic energy of emitted electrons for the two sources. [4]

15. A metal surface has a work function of 3.00 eV. Light of wavelength 350 nm is incident on the surface.

(a) Calculate the energy of a single photon of this light, in electronvolts. [2]

(b) Determine the maximum speed of the emitted electrons. [3]

Section D: Application and Analysis (Questions 16-20, 10 marks)

Answer all questions in this section.

16. Explain why, in the photoelectric effect, increasing the intensity of incident light increases the photoelectric current but does not affect the maximum kinetic energy of the emitted electrons. [2]

17. A particular metal has a threshold wavelength of 550 nm. Determine whether photoelectrons will be emitted when light of wavelength 600 nm is used. Explain your reasoning. [2]

18. In a photoelectric experiment, the maximum kinetic energy of emitted electrons is found to be 1.50 eV when light of frequency 9.00 × 10¹⁴ Hz is used. Calculate the work function of the metal, in joules. [2]

19. A student suggests that increasing the intensity of light below the threshold frequency could eventually cause electron emission if the light is shone for a long enough time. Explain why this suggestion is incorrect, using the photon model of light. [2]

20. Light of frequency 1.20 × 10¹⁵ Hz is incident on a metal surface with work function 4.00 eV. Calculate the de Broglie wavelength of the fastest emitted electrons. [2]

END OF QUIZ

Check your answers carefully before submitting.

Answers

A-Level Physics H1 Quiz - Modern Physics - ANSWER KEY

Total Marks: 50

Section A: Short Answer and Structured Response (Questions 1-5, 10 marks)

1. State what is meant by the photoelectric effect. [2]

Answer:

The photoelectric effect is the emission of electrons [B1] from a metal surface when electromagnetic radiation (light) of sufficiently high frequency is incident on it [B1].

Marking:

[B1] for "emission of electrons" or equivalent
[B1] for "from a metal surface when light/EM radiation is incident" with reference to frequency requirement
Accept: "Ejection of photoelectrons from a metal when light shines on it"

2. Define the term "work function" of a metal. [2]

Answer:

The work function is the minimum energy required [B1] to remove an electron from the surface of a metal [B1].

Marking:

[B1] for "minimum energy required"
[B1] for "to remove/emit an electron from the metal surface"
Accept: "The minimum energy needed for an electron to escape from the metal surface"

3. Explain why the photoelectric effect cannot be explained by the wave model of light. [3]

Answer:

According to the wave model, the energy of light depends on its intensity, not its frequency [B1].
The wave model predicts that electrons would be emitted after a time delay as energy accumulates, regardless of frequency [B1].
However, experiments show that emission is instantaneous (no time delay) and only occurs above a threshold frequency, regardless of intensity [B1].

Marking:

[B1] for stating wave model prediction about energy depending on intensity
[B1] for stating wave model prediction about time delay/energy accumulation
[B1] for contrasting with experimental observations (instantaneous emission and/or threshold frequency)
Accept any two valid contrasts with experimental evidence

4. State two observations from photoelectric effect experiments that support the photon model of light. [2]

Answer: Any two from:

Electrons are emitted only when the frequency of light exceeds a threshold frequency [B1]
Electron emission is instantaneous (no time delay) [B1]
Maximum kinetic energy of electrons depends on frequency, not intensity [B1]
Increasing intensity increases the number of electrons emitted but not their maximum kinetic energy [B1]

Marking:

[B1] for each valid observation (maximum 2 marks)

5. A metal surface has a work function of 2.30 eV. Calculate the threshold frequency for this metal. [3]

Answer:

Φ = hf₀
Φ = 2.30 eV = 2.30 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J [M1]
f₀ = Φ / h = 3.68 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ [M1]
f₀ = 5.55 × 10¹⁴ Hz [A1]

Marking:

[M1] for converting eV to joules
[M1] for correct substitution into f₀ = Φ/h
[A1] for correct answer with units: 5.55 × 10¹⁴ Hz (accept 5.5 × 10¹⁴ Hz)

Section B: Core Concepts and Equations (Questions 6-10, 10 marks)

6. Explain what is meant by "stopping potential" in the context of the photoelectric effect. [2]

Answer:

The stopping potential is the minimum potential difference (voltage) [B1] that must be applied to stop the most energetic photoelectrons from reaching the collector/anode [B1].

Marking:

[B1] for "minimum potential difference/voltage"
[B1] for "to stop the most energetic/fastest electrons" or "to reduce photocurrent to zero"
Accept: "The retarding potential at which the photoelectric current becomes zero"

7. State Einstein's photoelectric equation and define each symbol used. [3]

Answer:

hf = Φ + K.E._max [B1]
h = Planck's constant [B1]
f = frequency of incident light
Φ = work function of the metal
K.E._max = maximum kinetic energy of emitted electrons [B1]

Marking:

[B1] for correct equation (accept hf = Φ + ½mv²_max or hf = Φ + eV_s)
[B1] for identifying h and f
[B1] for identifying Φ and K.E._max (or equivalent)
Award full marks if all symbols are correctly defined alongside the equation

8. Ultraviolet light of wavelength 200 nm is incident on a metal surface with work function 4.50 eV. Determine whether electrons will be emitted from the surface. Show your working. [3]

Answer:

Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (200 × 10⁻⁹) [M1]
E = 9.945 × 10⁻¹⁹ J [M1]
E = 9.945 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 6.22 eV [M1]
Since 6.22 eV > 4.50 eV (photon energy exceeds work function), electrons WILL be emitted [A1]

Marking:

[M1] for using E = hc/λ
[M1] for correct calculation of photon energy (in J or eV)
[A1] for correct comparison and conclusion
Accept alternative method: calculate threshold wavelength and compare

9. A clean metal surface is illuminated with light of wavelength 5.50 × 10⁻⁷ m. The stopping potential is measured to be 0.38 V. Calculate the work function of the metal, in electronvolts. [2]

Answer:

Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (5.50 × 10⁻⁷) = 3.616 × 10⁻¹⁹ J [M1]
E = 3.616 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.26 eV
K.E._max = eV_s = 0.38 eV
Φ = E - K.E._max = 2.26 - 0.38 = 1.88 eV [A1]

Marking:

[M1] for calculating photon energy and using Φ = hf - eV_s (or equivalent)
[A1] for correct answer: 1.88 eV (accept 1.9 eV)

Answer:

Φ = hf₀ [M1]
For f = 2f₀: K = h(2f₀) - hf₀ = hf₀
For f = 3f₀: K.E._max = h(3f₀) - hf₀ = 2hf₀ = 2K [A1]

Marking:

[M1] for relating Φ to f₀ and setting up equations
[A1] for correct answer: 2K

Section C: Calculations and Data Analysis (Questions 11-15, 20 marks)

11. Monochromatic light of frequency 8.00 × 10¹⁴ Hz is incident on a metal surface with work function 2.00 eV.

(a) Calculate the energy of a single photon of this light, in joules. [2]

Answer:

E = hf = 6.63 × 10⁻³⁴ × 8.00 × 10¹⁴ [M1]
E = 5.30 × 10⁻¹⁹ J [A1]

Marking:

[M1] for correct formula and substitution
[A1] for correct answer: 5.30 × 10⁻¹⁹ J

(b) Calculate the maximum kinetic energy of the emitted electrons, in joules. [2]

Answer:

Φ = 2.00 eV = 2.00 × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹⁹ J [M1]
K.E._max = hf - Φ = 5.30 × 10⁻¹⁹ - 3.20 × 10⁻¹⁹ = 2.10 × 10⁻¹⁹ J [A1]

Marking:

[M1] for converting work function to joules and applying K.E._max = hf - Φ
[A1] for correct answer: 2.10 × 10⁻¹⁹ J

(c) Calculate the stopping potential required to prevent electron emission. [2]

Answer:

eV_s = K.E._max [M1]
V_s = K.E._max / e = 2.10 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.31 V [A1]

Marking:

[M1] for using eV_s = K.E._max
[A1] for correct answer: 1.31 V (accept 1.3 V)

12. Photoelectric experiment data analysis.

(a) Calculate the frequency corresponding to each wavelength. Tabulate your results. [3]

Answer: Using f = c/λ:

λ / nm	λ / m	f / Hz	E_max / eV
450	4.50 × 10⁻⁷	6.67 × 10¹⁴	0.55
400	4.00 × 10⁻⁷	7.50 × 10¹⁴	0.90
350	3.50 × 10⁻⁷	8.57 × 10¹⁴	1.35
300	3.00 × 10⁻⁷	1.00 × 10¹⁵	1.95
250	2.50 × 10⁻⁷	1.20 × 10¹⁵	2.75

Marking:

[M1] for using f = c/λ correctly
[M1] for correct conversion of nm to m
[A1] for all five frequencies correct (allow ±0.02 × 10¹⁴ Hz)
Award 2 marks if 3-4 frequencies are correct

(b) Plot a graph of E_max (y-axis) against frequency f (x-axis) on the grid provided. [4]

Answer:

Correct axes with labels and units: E_max / eV on y-axis, f / 10¹⁴ Hz on x-axis [B1]
Appropriate scales using more than half the grid [B1]
All five points plotted correctly [B1]
Best-fit straight line drawn [B1]

Marking:

[B1] for correct axes labels and units
[B1] for appropriate linear scales
[B1] for correct plotting of points (±½ small square)
[B1] for reasonable best-fit straight line

(c) Use your graph to determine the value of Planck's constant. [3]

Answer:

Gradient = ΔE_max / Δf [M1]
From graph: gradient ≈ (2.75 - 0.55) / (1.20 × 10¹⁵ - 6.67 × 10¹⁴) = 2.20 / (5.33 × 10¹⁴) = 4.13 × 10⁻¹⁵ eV·s [M1]
h = gradient × e = 4.13 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.61 × 10⁻³⁴ J·s [A1]

Marking:

[M1] for identifying gradient = h/e or using ΔE_max/Δf
[M1] for correct calculation of gradient from graph
[A1] for h in range 6.0-7.0 × 10⁻³⁴ J·s (accept 6.6 × 10⁻³⁴ J·s)
Note: Exact value depends on student's graph; accept values within reasonable range

(d) Use your graph to determine the work function of the metal. [2]

Answer:

From E_max = hf - Φ, the x-intercept gives threshold frequency f₀ [M1]
Φ = hf₀ (or read y-intercept magnitude and convert)
From graph: f₀ ≈ 5.8 × 10¹⁴ Hz, so Φ = 6.63 × 10⁻³⁴ × 5.8 × 10¹⁴ = 3.85 × 10⁻¹⁹ J = 2.4 eV [A1]

Marking:

[M1] for correct method (x-intercept or y-intercept approach)
[A1] for work function in range 2.2-2.6 eV (or equivalent in joules)
Accept answer derived from student's own graph

13. Stopping potential vs frequency graph analysis.

(a) State the physical quantity represented by the gradient of this graph. [1]

Answer:

h/e (Planck's constant divided by elementary charge) [B1]

Marking:

[B1] for h/e or "Planck's constant / e"

(b) State the physical quantity represented by the x-intercept of this graph. [1]

Answer:

Threshold frequency (f₀) [B1]

Marking:

[B1] for "threshold frequency" or f₀

(c) Explain why the graph is a straight line and does not pass through the origin. [2]

Answer:

From eV_s = hf - Φ, rearranging gives V_s = (h/e)f - Φ/e, which is of the form y = mx + c (straight line) [B1]
The graph does not pass through the origin because there is a non-zero y-intercept (-Φ/e), meaning a minimum frequency (threshold frequency) is required before electron emission occurs [B1]

Marking:

[B1] for linking to straight line equation y = mx + c
[B1] for explaining non-zero intercept due to work function/threshold frequency

Compare and explain the photoelectric current and the maximum kinetic energy of emitted electrons for the two sources. [4]

Answer:

Photoelectric current: Source A will produce twice the photoelectric current of Source B [B1]. This is because intensity is proportional to the number of photons incident per unit time; twice the intensity means twice the number of photons, leading to twice the number of emitted electrons per unit time [B1].
Maximum kinetic energy: The maximum kinetic energy of emitted electrons will be the same for both sources [B1]. This is because K.E._max = hf - Φ, and since frequency f is the same for both sources and Φ is constant for the metal, K.E._max is unchanged [B1].

Marking:

[B1] for stating current is greater for Source A (or twice)
[B1] for explaining in terms of photon number/electron emission rate
[B1] for stating K.E._max is the same
[B1] for explaining using Einstein's equation (depends on frequency, not intensity)

15. A metal surface has a work function of 3.00 eV. Light of wavelength 350 nm is incident on the surface.

(a) Calculate the energy of a single photon of this light, in electronvolts. [2]

Answer:

E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (350 × 10⁻⁹) = 5.683 × 10⁻¹⁹ J [M1]
E = 5.683 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.55 eV [A1]

Marking:

[M1] for correct calculation of photon energy in joules
[A1] for correct conversion to eV: 3.55 eV

(b) Determine the maximum speed of the emitted electrons. [3]

Answer:

K.E._max = E - Φ = 3.55 eV - 3.00 eV = 0.55 eV [M1]
K.E._max = 0.55 × 1.60 × 10⁻¹⁹ = 8.80 × 10⁻²⁰ J [M1]
½mv² = K.E._max → v = √(2 × 8.80 × 10⁻²⁰ / 9.11 × 10⁻³¹) = √(1.932 × 10¹¹) = 4.39 × 10⁵ m s⁻¹ [A1]

Marking:

[M1] for calculating K.E._max in eV or J
[M1] for using ½mv² = K.E._max
[A1] for correct answer: 4.4 × 10⁵ m s⁻¹ (accept 4.39 × 10⁵ m s⁻¹)

Section D: Application and Analysis (Questions 16-20, 10 marks)

Answer:

Increasing intensity increases the number of photons incident per unit time [B1], which ejects more electrons per unit time, increasing the current.
The maximum kinetic energy depends only on photon frequency (K.E._max = hf - Φ), not on the number of photons [B1].

Marking:

[B1] for linking intensity to photon number/electron emission rate
[B1] for stating K.E._max depends on frequency, not intensity

17. A particular metal has a threshold wavelength of 550 nm. Determine whether photoelectrons will be emitted when light of wavelength 600 nm is used. Explain your reasoning. [2]

Answer:

No, photoelectrons will not be emitted [B1].
The threshold wavelength is the maximum wavelength that can cause emission. Since 600 nm > 550 nm, the photon energy (E = hc/λ) is less than the work function, so emission cannot occur [B1].

Marking:

[B1] for correct conclusion (no emission)
[B1] for correct reasoning (longer wavelength means lower energy, below threshold)

Answer:

hf = 6.63 × 10⁻³⁴ × 9.00 × 10¹⁴ = 5.967 × 10⁻¹⁹ J [M1]
K.E._max = 1.50 eV = 1.50 × 1.60 × 10⁻¹⁹ = 2.40 × 10⁻¹⁹ J
Φ = hf - K.E._max = 5.967 × 10⁻¹⁹ - 2.40 × 10⁻¹⁹ = 3.57 × 10⁻¹⁹ J [A1]

Marking:

[M1] for calculating photon energy and converting K.E._max to joules
[A1] for correct answer: 3.57 × 10⁻¹⁹ J (accept 3.6 × 10⁻¹⁹ J)

Answer:

In the photon model, each photon interacts with a single electron [B1]. If the photon energy (hf) is less than the work function (Φ), the electron cannot absorb enough energy to escape, regardless of how many photons arrive (intensity) or for how long.
Energy from multiple photons cannot be accumulated by a single electron to overcome the work function [B1].

Marking:

[B1] for stating one-to-one photon-electron interaction
[B1] for explaining that energy cannot be accumulated from multiple photons

20. Light of frequency 1.20 × 10¹⁵ Hz is incident on a metal surface with work function 4.00 eV. Calculate the de Broglie wavelength of the fastest emitted electrons. [2]

Answer:

Photon energy E = hf = 6.63 × 10⁻³⁴ × 1.20 × 10¹⁵ = 7.956 × 10⁻¹⁹ J = 4.97 eV
K.E._max = E - Φ = 4.97 eV - 4.00 eV = 0.97 eV = 1.552 × 10⁻¹⁹ J [M1]
v = √(2K.E._max / m) = √(2 × 1.552 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = 5.84 × 10⁵ m s⁻¹
λ = h / mv = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 5.84 × 10⁵) = 1.25 × 10⁻⁹ m [A1]

Marking:

[M1] for calculating K.E._max and electron speed or momentum
[A1] for correct de Broglie wavelength: 1.25 × 10⁻⁹ m (accept 1.2-1.3 × 10⁻⁹ m)

END OF ANSWER KEY