CAIE A-Level Physics A2 22.1 Energy and Momentum of a Photon Questions

Practise analysing photon quanta, calculating energy, wavelength, momentum and radiation force, and using power to find photon emission rate.

Exam points

distinguish particulate nature and photon quanta in electromagnetic radiation
calculate photon energy with E = hf or E = hc/? and convert between joules and electronvolts
use photon energy and power to determine photon emission rate
calculate photon momentum and radiation force using p = E/c and momentum change

Covered syllabus points

22.1.1—Electromagnetic radiation has a particulate nature 22.1.2—A photon is a quantum of electromagnetic energy 22.1.3—E = hf 22.1.4—The electronvolt (eV) as a unit of energy 22.1.5—A photon has momentum and that the momentum is given by p = E / c

[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 $4.1 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$ .

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

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question 2

Question 3