# Exam-style Questions: Quantum Physics

1. a) In an experiment, an electrically insulated zinc plate is negatively charged. When exposed to a weak ultra-violet source, the plate starts to lose its negative charge.

Explain this phenomenon in terms of the photoelectric effect. Suggest how increasing the intensity of the ultra-violet source would affect the experiment.

(You will be awarded marks for the quality of written communication in your answer to this question.)

b) A 1.0W laser produces red light of wavelength 6.3 x 10-7m.

Calculate

i) the frequency of the radiation,

frequency = .............................unit...................

ii) the energy of a photon of red light.

energy =..........................................J

iii) Calculate the number of photons emitted per second by the laser.

Number =.....................................s-1

iv) State how, and explain why, the number of photons emitted per second would change if the 1.0mW laser produced blue light.

(Marks available: 16)

Answer outline and marking scheme for question: 1

a) Any five from:

• photons are involved / E=hf
• surface electrons are involved
• one to one exchange of energy between photon and electron
• electrons carry negative charge, therefore removal means resuction in (negative) charge (of plate)
• electron released when photon energy > workfunction (energy)
• electron emission related to threshold frequency
• energy conserved in the interaction (between photon and electron).
• Einstein's equation mentioned (hf = 1/2 mv2 + ø)

(5 marks)

Any two from:

• More photons (in a given time)
• More electrons are removed (in a given time)
• The plate loses charge quicker.

(2 marks)

b) i) f = 3.0 x 108 / 6.3 x 10-7

f = 4.7 (6) x 1014

unit: hertz, Hz

(3 marks)

ii) E = hf

E = 6.63 x 10-34 x 4.76 x 1014 = 3.1(6) x 10-19 (J)

(2 marks)

iii) N = 1 x 10-3/ 3.16 x 10-19

N = 3.1(7) x 1015 (s-1)

(2 marks)

iv) Energy of photon is greater / blue light has shorter wavelength

Reduced (rate) of photons / fewer photons ( in a given time)

(2 marks)

(Marks available: 16)

2. Fig. 7.1 shows a photocell.

When the metal surface is exposed to electromagnetic radiation, photoelctrons are ejected. The collector collects the photoelectrons and the sensitive ammeter indicates the presence of a tiny current.

a) For a certain frequency and intensity of radiation, the ammeter shows a current of 1.2 x 10-7 A. Calculate:

i) the charge reaching the collector in 5.0 s

charge = ............................. C

ii) the number of photoelectrons reaching the collector in 5.0 s.

number of electrons = ..........................

(3 Marks)

b) The work function energy of the metal is 3.5 x 10-19 J and the incident radiation has frequency 7.0 x 1014 Hz. Calculate the maximum kinetic energy of an ejected photoelectron.

energy = ....................................... J

(3 Marks)

c) The intensity of the incident radiation is doubled but the wavelength is kept constant. State the effect this has on each of the following

i) the energy of each photon

(1 Mark)

ii) the maximum kinetic energy of each photoelectron

(1 Mark)

iii) the currrent in the photocell

(1 Mark)

(Marks available: 9)

Answer outline and marking scheme for question: 2

a) i) ΔQ = I Δt / 1.2 x 10-7 x 5 (Any subject and no need for Δ notation)

charge = 6.0 x 10-7 (C) (Allow 6 x 10-7 (C))

ii)

 number = 6.0 x 10-7 3.75 x 1012 ≈ 3.8 x 1012 (Possible ECF) 1.6 x 10-19

(3 Marks)

b) E = hf / E = 6.63 x 10-34 x 7.0 x 1014 = 4.64 x 10-19≈ 4.6 x 10-19 (J)

hf = φ + KE max (Allow other subject for photoelectric equation)

KEmax = (4.64 - 3.5) x 10-19= 1.14 x 10-19≈1.1 x 10-19 (J)

(3 Marks)

c) i)Energy of photon is the same

ii) KE of electron is the same

iii) The current doubled (as there are twice as many photons)

(1 Mark)

(Marks available: 9)