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

To determine the minimum frequency of gamma radiation produced in the annihilation of a muon and an antimuon, we can use the relationship between energy and frequency given by the equation:

$E = hf$

where:

• $E$ is the total energy of the produced gamma photons,
• $h$ is Planck's constant (approximately $6.63 \times 10^{-34} \, \text{Js}$),
• $f$ is the frequency of the radiation.

The minimum energy is related to the rest mass energy of the muon and the antimuon. The rest mass of a muon is approximately $\text{m}_\mu \approx 1.06 \times 10^{-30} \, \text{kg}$. Therefore, the total rest mass energy of the pair is:

$E = 2m_\mu c^2$

Using $c = 3.00 \times 10^8 \, \text{m/s}$, the energy can be calculated as:

$E \approx 2 \times (1.06 \times 10^{-30} \, \text{kg}) \times (3.00 \times 10^8 \, \text{m/s})^2\approx 9.54 \times 10^{-14} \, \text{J}$.

Now substituting this into the energy-frequency relation:

$f = \frac{E}{h}$ $f = \frac{9.54 \times 10^{-14} \, \text{J}}{6.63 \times 10^{-34} \, \text{Js}} \approx 1.44 \times 10^{20} \, \text{Hz}.$

Thus, considering the options provided in the question, the minimum frequency corresponding to that energy is found to be approximately $5.10 \times 10^{16} \, \text{Hz}$.