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A 1 MeV positron and a 1 MeV electron meet each moving in opposite directions. They annihilate each other by emitting two photons. If the rest mass energy of an electron is 0.51 MeV, the wavelength of each photon is?

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Last updated date: 13th Jun 2024
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Answer
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Hint: The phenomenon when an electron collides with the positron at low energy and then the particles annihilates into two $\gamma $ ray photons is known as pair-annihilation. The electron and positron have the same mass but the charge on the electron is negative and charge on the positron is positive and the magnitude of the electron and positron are equal.

Formula used:
The formula of the energy of photon is given by,
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$
Where energy is E, the Planck’s constant is h, the speed of light is c and the wavelength is $\lambda $.

Complete step by step solution:
It is given in the problem that 1 MeV positron and a 1 MeV electron meet each moving in opposite directions they annihilate each other by emitting two photons the rest mass energy of an electron is 0.51 MeV and we need to find the wavelength of each photon.
The equation involved in the annihilation is equal to,
$ \Rightarrow {e^ + } + {e^ - } = \gamma + \gamma $
The total energy of the positron is equal to,
$ \Rightarrow 1 + 0.512 = 1.512MeV$
The total energy of the electron is equal to,
$ \Rightarrow 1 + 0.512 = 1.512MeV$
The total energy of positron and electron is equal to,
$ \Rightarrow 2 \times 1.512 = 3.024MeV$
Since this energy is distributed in two photons and therefore the energy of each photon is equal to,
$ \Rightarrow E = \dfrac{{3.024}}{2} = 1.512MeV$
The formula of the energy of photon is given by,
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$
Where energy is E, the Planck’s constant is h, the speed of light is c and the wavelength is $\lambda $.
The energy of A photon is$E = 1.512MeV$, the value of Planck’s constant is $h = 6.62 \times {10^{ - 34}}J - s$ and speed of light is equal to $c = 3 \times {10^8}\dfrac{m}{s}$.
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$
$ \Rightarrow \lambda = \dfrac{{hc}}{E}$
$ \Rightarrow \lambda = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.512 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}$
$ \Rightarrow \lambda = \dfrac{{19.86 \times {{10}^{8 - 34}}}}{{2.419 \times {{10}^{6 - 19}}}}$
$ \Rightarrow \lambda = \dfrac{{8.21 \times {{10}^{ - 26}}}}{{{{10}^{ - 13}}}}$
$ \Rightarrow \lambda = 8.21 \times {10^{ - 26 + 13}}$
$ \Rightarrow \lambda = 8.21 \times {10^{ - 13}}m$.

The wavelength of the photon is equal to $\lambda = 8.21 \times {10^{ - 13}}m$.

Note: The students are advised to understand and remember the formula of the energy of the photons. In order to solve this problem, we need to convert the energy from mega electron volts to joules. The mega electron volts in joules is equal to $1MeV = 1.6 \times {10^{ - 19}}J$.