Two gamma photons are produced when a muon and an antimuon annihilate each other - AQA - A-Level Physics - Question 10 - 2018 - Paper 1

To find the minimum frequency of the gamma radiation produced, we will use the equation that relates energy to frequency:

$E = hf$

where:

• $E$ is the energy in joules,
• $h$ is Planck's constant ($6.626 \times 10^{-34} \text{ J s}$), and
• $f$ is the frequency in hertz.

First, we need to convert the energy from MeV to joules:

$1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}$ Thus,

$E = 211.4 \text{ MeV} \times 1.602 \times 10^{-13} \text{ J/MeV} \approx 3.39 \times 10^{-11} \text{ J}$

Now, we can solve for frequency:

$f = \frac{E}{h} = \frac{3.39 \times 10^{-11} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} \approx 5.11 \times 10^{22} \text{ Hz}$

Based on the calculations, the minimum frequency of the gamma radiation produced is approximately $5.11 \times 10^{22} \text{ Hz}$, which corresponds to option D.