CAIE A-Level Physics A2 22 Quantum Physics Questions

Practise analysing photon energy and momentum, photoelectric emission, wave-particle evidence, de Broglie matter waves and quantised line spectra.

Exam points

calculate and interpret photon energy and momentum, including energy, frequency, wavelength and radiation-force relationships
explain and calculate photoelectric emission using threshold conditions, work function and maximum kinetic energy
use photoelectric, interference and diffraction evidence to evaluate wave�particle duality and apply de Broglie wavelength
interpret atomic energy levels and emission or absorption spectra as quantised transitions
calculate photon frequency or energy from an atomic energy-level difference

Covered syllabus points

22.1 Energy and momentum of a photon 22.2 Photoelectric effect 22.3 Wave–particle duality 22.4 Energy levels in atoms and line spectra

[Maximum number: 7]

(a)

State what is meant by a photon.

[ 2 ]

(b)

For an accelerating voltage of 32 kV in Fig. 8.1, determine:

[ 5 ]

(i)

the maximum momentum of an X-ray photon produced at the target

maximum photon momentum =

[ 2 ]

momentum =E / c
momentum $=\left(0.032 \times 1.60 \times 10^{-13}\right) /\left(3.00 \times 10^{8}\right)$ $=1.7 \times 10^{-23} \mathrm{~N} \mathrm{~s}$

(ii)

the minimum wavelength of X-rays produced at the target.

minimum wavelength =m [3]

[ 3 ]

E=h f and $\lambda=c / f$ $\begin{aligned} \lambda =h c / E =\left(6.63 \times 10^{-34} \times 3.00 \times 10^{8}\right) /\left(0.032 \times 1.60 \times 10^{-13}\right) \end{aligned}$ $\lambda=3.9 \times 10^{-11} \mathrm{~m}$

[Maximum number: 2]

Fig． 8.1 shows part of the emission spectrum of visible radiation emitted by hydrogen gas in a star in a distant galaxy．

Fig． 8.1

The galaxy is moving away from the Earth at a speed of $6.2 \times 10^{6} \mathrm{~ms}^{-1}$ ．

Calculate the energy of a photon of this radiation．

photon energy =J [2]

E=h f and $\lambda=c / f$ $\begin{aligned} E =\left(6.63 \times 10^{-34} \times 3.00 \times 10^{8}\right) /\left(488 \times 10^{-9}\right) =4.08 \times 10^{-19} \mathrm{~J} \end{aligned}$

[Maximum number: 2]

Electrons are accelerated from rest through a potential difference (p.d.) of 4.8 kV .

Show that the final speed of the electrons is $4.1 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$ .

$\frac{1}{2} m v^{2}=e V$ $1 / 2 \times 9.11 \times 10^{-31} \times v^{2}=1.60 \times 10^{-19} \times 4800 \mathrm{so} v=4.1 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question 2

Question 3