Physics questions, chapter 7

Nama / NameKelas / ClassTarikh / Date

Quantum Physics

46 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first. Use h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C and electron mass 9.11 × 10⁻³¹ kg.

Section A: objective questions

13 marks

Choose the best answer for each question.

Question 1

What is a black body?

A body that is black in colour and emits no radiation
An idealised body that absorbs all the electromagnetic radiation falling on it
A body that reflects all radiation falling on it
A body that emits only visible light

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A black body absorbs all radiation that falls on it and also emits thermal radiation that depends on its temperature.

For the teacher: Remembering, low difficulty, standard 7.1.1

Textbook p. 224

Question 2

A black body is heated from 4 000 K to 7 000 K. Which change happens to the radiation graph?

The peak moves to a longer wavelength
The peak stays at the same wavelength
The peak moves to a shorter wavelength
The total radiation intensity decreases

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As temperature rises, the wavelength at maximum intensity becomes shorter and the intensity increases.

For the teacher: Understanding, medium difficulty, standard 7.1.1

Textbook p. 225

Question 3

Why can a line spectrum be used to identify an element?

Each element produces its own distinctive set of lines
All elements produce the same lines
The lines are continuous with no gaps
The lines depend on the temperature only

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The wavelengths of the lines are unique to each element.

For the teacher: Understanding, low difficulty, standard 7.1.2

Textbook p. 228

Question 4

The frequency of a light wave is doubled. What happens to the energy of each photon?

It is halved
It stays the same
It is multiplied by four
It is doubled

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E = hf, so photon energy is directly proportional to frequency.

For the teacher: Understanding, low difficulty, standard 7.1.2

Textbook p. 229

Question 5

What is the energy of a photon of light with a wavelength of 500 nm?

1.33 × 10⁻²⁷ J
3.98 × 10⁻¹⁹ J
1.99 × 10⁻²⁵ J
3.98 × 10⁻²⁸ J

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E = hc ÷ λ = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (500 × 10⁻⁹) = 3.98 × 10⁻¹⁹ J. Remember to convert nm to m.

For the teacher: Applying, medium difficulty, standard 7.1.4

Textbook p. 229

Question 6

Four particles move with the same speed. Which has the shortest de Broglie wavelength?

Electron
Proton
Neutron
Alpha particle

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λ = h ÷ mv. For the same speed the particle with the greatest mass has the shortest wavelength, which is the alpha particle.

For the teacher: Understanding, medium difficulty, standard 7.1.3

Textbook p. 230

Question 7

The intensity of light shining on a metal above its threshold frequency is increased. What happens?

The kinetic energy of each photoelectron increases
The photoelectrons are emitted after a longer delay
The number of photoelectrons emitted per second increases
The threshold frequency decreases

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More photons arrive each second, so more photoelectrons are emitted. Kinetic energy depends on frequency, not intensity.

For the teacher: Understanding, low difficulty, standard 7.2.2

Textbook p. 237

Question 8

A metal has a threshold frequency of 6.0 × 10¹⁴ Hz. Which light, in a vacuum, will cause photoelectric emission?

Wavelength 450 nm
Wavelength 520 nm
Wavelength 600 nm
Wavelength 700 nm

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The threshold wavelength is λ₀ = c ÷ f₀ = 3.00 × 10⁸ ÷ 6.0 × 10¹⁴ = 500 nm. Only a wavelength shorter than 500 nm (frequency above f₀) works.

For the teacher: Applying, medium difficulty, standard 7.3.2

Textbook p. 239

Question 9

A metal has a threshold frequency of 5.0 × 10¹⁴ Hz. What is its work function?

1.33 × 10⁻⁴⁸ J
1.66 × 10⁻¹⁹ J
3.32 × 10⁻¹⁹ J
3.32 × 10⁻²⁵ J

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W = hf₀ = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.32 × 10⁻¹⁹ J.

For the teacher: Applying, medium difficulty, standard 7.3.2

Textbook p. 239

Question 10

Diagram 1 shows graphs of the maximum kinetic energy Kmax against frequency f for metals A and B. Which statement is correct?

Two parallel lines of Kmax against f; the line for B meets the frequency axis at a higher frequency than the line for A — Diagram 1

Metal A has the greater work function
Metal B has the greater work function and the gradient of both graphs is h
Metal A has the greater gradient because its threshold frequency is lower
Both metals have the same work function because the gradients are equal

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W = hf₀ and f₀ of B is greater. The gradient of Kmax against f is h for every metal, so the lines are parallel.

For the teacher: Analysing, high difficulty, standard 7.3.3

Textbook p. 239

Question 11

Light of frequency 7.5 × 10¹⁴ Hz shines on a metal of work function 3.0 × 10⁻¹⁹ J. What is the maximum kinetic energy of a photoelectron?

3.00 × 10⁻¹⁹ J
4.97 × 10⁻¹⁹ J
7.97 × 10⁻¹⁹ J
1.97 × 10⁻¹⁹ J

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hf = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.97 × 10⁻¹⁹ J. Kmax = hf − W = 4.97 × 10⁻¹⁹ − 3.0 × 10⁻¹⁹ = 1.97 × 10⁻¹⁹ J.

For the teacher: Applying, medium difficulty, standard 7.3.4

Textbook p. 241

Question 12

A lamp emits 2.0 W of light of wavelength 600 nm. How many photons are emitted per second?

6.03 × 10¹⁸
3.32 × 10¹⁸
1.66 × 10⁻¹⁹
1.99 × 10⁻²⁵

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n = Pλ ÷ hc = (2.0)(600 × 10⁻⁹) ÷ (6.63 × 10⁻³⁴ × 3.00 × 10⁸) = 6.03 × 10¹⁸ per second.

For the teacher: Applying, medium difficulty, standard 7.1.5

Textbook p. 232

Question 13

Which device works by the photoelectric effect?

Electric kettle
Loudspeaker
Light detector of an automatic door
Transformer

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The light beam and photocell act as a switch; breaking the beam stops the photoelectric current and the door opens.

For the teacher: Remembering, low difficulty, standard 7.3.6

Textbook p. 244

Section B: structured questions

33 marks

Answer all questions. Show your working.

Question 14

The violet line of a mercury vapour lamp has a frequency of 7.41 × 10¹⁴ Hz.

[2 marks] (a)Calculate the energy of one photon of this line.

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E = hf = 6.63 × 10⁻³⁴ × 7.41 × 10¹⁴ [1 mark]1
= 4.91 × 10⁻¹⁹ J [1 mark]1

Tick each point you got.

[2 marks] (b)Calculate its wavelength in nm.

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λ = c ÷ f = 3.00 × 10⁸ ÷ 7.41 × 10¹⁴ [1 mark]1
= 4.05 × 10⁻⁷ m = 405 nm [1 mark]1

Tick each point you got.

[1 mark] (c)Why can this line spectrum be used to show that mercury is present in a lamp?

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Every element gives its own unique set of line wavelengths1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 7.1.2

Textbook p. 228

Question 15

An electron of mass 9.11 × 10⁻³¹ kg moves at 2.0 × 10⁶ m s⁻¹. A ball of mass 0.15 kg moves at 20 m s⁻¹.

[2 marks] (a)Calculate the de Broglie wavelength of the electron.

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λ = h ÷ mv = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.0 × 10⁶) [1 mark]1
= 3.64 × 10⁻¹⁰ m [1 mark]1

Tick each point you got.

[2 marks] (b)Calculate the de Broglie wavelength of the ball.

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λ = 6.63 × 10⁻³⁴ ÷ (0.15 × 20) [1 mark]1
= 2.21 × 10⁻³⁴ m [1 mark]1

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[1 mark] (c)Why can wave behaviour not be observed for the ball?

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Its wavelength is far too short to be detected (Planck's constant is very small)1

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For the teacher: Applying, medium difficulty, standard 7.1.3

Textbook p. 230

Question 16

A green laser pointer emits light of wavelength 532 nm with an output power of 1.0 mW.

[2 marks] (a)Calculate the energy of one photon.

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E = hc ÷ λ = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (532 × 10⁻⁹) [1 mark]1
= 3.74 × 10⁻¹⁹ J [1 mark]1

Tick each point you got.

[2 marks] (b)Calculate the number of photons emitted per second.

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n = P ÷ E = 1.0 × 10⁻³ ÷ 3.74 × 10⁻¹⁹ [1 mark]1
= 2.67 × 10¹⁵ s⁻¹ [1 mark]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 7.1.5

Textbook p. 232

Question 17

A metal has a threshold frequency of 5.5 × 10¹⁴ Hz. Violet light (7.5 × 10¹⁴ Hz) and a bright red light (4.3 × 10¹⁴ Hz) are shone separately on the metal.

[1 mark] (a)Which light causes photoelectric emission?

AViolet lightBRed light

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Violet light

Only its frequency is above the threshold frequency.

[2 marks] (b)Explain why the bright red light does not emit photoelectrons.

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Each photon of red light has energy hf less than the work function [1 mark]1
An electron absorbs one photon at a time, so brightness (more photons) does not help [1 mark]1

Tick each point you got.

[2 marks] (c)The intensity of the violet light is increased. State what happens to the number of photoelectrons per second and to their maximum kinetic energy.

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The number of photoelectrons per second increases [1 mark]1
The maximum kinetic energy stays the same [1 mark]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 7.2.2

Textbook p. 237

Question 18

The work function of sodium is 2.28 eV. Use hc = 1.243 × 10³ eV nm, 1 eV = 1.60 × 10⁻¹⁹ J and electron mass 9.11 × 10⁻³¹ kg.

[2 marks] (a)Will red light of wavelength 680 nm cause photoelectric emission from sodium? Show by calculation.

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Photon energy = 1243 ÷ 680 = 1.83 eV [1 mark]1
1.83 eV is less than 2.28 eV, so no emission [1 mark]1

Tick each point you got.

[2 marks] (b)Calculate the threshold wavelength of sodium.

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λ₀ = hc ÷ W = 1243 ÷ 2.28 [1 mark]1
= 545 nm [1 mark]1

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[2 marks] (c)Violet light of wavelength 400 nm shines on sodium. Calculate the maximum kinetic energy of a photoelectron in joules.

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Photon energy = 1243 ÷ 400 = 3.11 eV; Kmax = 3.11 − 2.28 = 0.83 eV [1 mark]1
= 0.83 × 1.60 × 10⁻¹⁹ = 1.32 × 10⁻¹⁹ J [1 mark]1

Tick each point you got.

For the teacher: Applying, high difficulty, standard 7.3.4

Textbook p. 241

Question 19

Diagram 2 shows the graph of the maximum kinetic energy Kmax of photoelectrons against the frequency f of light for a metal.

Straight line of Kmax in eV against f in units of 10 to the 14 Hz, meeting the axis at 5.0 and passing through the point 8.0, 1.24 — Diagram 2

[1 mark] (a)State the threshold frequency of the metal.

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5.0 × 10¹⁴ Hz1

Tick each point you got.

[3 marks] (b)Calculate the gradient of the graph in J s. State what it represents.

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Gradient = 1.24 ÷ (3.0 × 10¹⁴) = 4.13 × 10⁻¹⁵ eV s [1 mark]1
× 1.60 × 10⁻¹⁹ = 6.6 × 10⁻³⁴ J s [1 mark]1
Planck's constant [1 mark]1

Tick each point you got.

[1 mark] (c)Calculate the work function of the metal in eV.

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W = hf₀ = 4.13 × 10⁻¹⁵ × 5.0 × 10¹⁴ = 2.07 eV1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 7.3.3

Textbook p. 239

Question 20

Diagram 3 shows the radiation from black bodies at 4 000 K (which looks red), 5 270 K (white) and 7 000 K (blue).

Radiation intensity against wavelength for black bodies at 4 000 K, 5 270 K and 7 000 K; the peak moves to shorter wavelength as temperature rises — Diagram 3: black-body curves at different temperatures

[2 marks] (b)Describe how the wavelength of maximum intensity and the colour change as a black body is heated from 4 000 K to 7 000 K.

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The wavelength of maximum intensity becomes shorter [1 mark]1
The colour changes from red to white and then to blue [1 mark]1

Tick each point you got.

[1 mark] (c)Classical physics predicted that the intensity would increase without limit at short wavelength. What do the curves show?

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The intensity falls at short wavelengths after the peak, so classical theory fails1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 7.1.1

Textbook p. 225

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