Question

# A positron and an electron can annihilate each other on colliding, producing energy as photons:${_{-1}^{0} e}+{_{+1}^{0} e} \longrightarrow 2_{0}^{0} \gamma$Assuming that both $\gamma$ rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

Solution

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We are given the reaction:

\begin{align*} ^0_{-1}\text{e + }^0_{+1}\text{e}\rightarrow\text{ 2 }^0_0\gamma \end{align*}

Calculate the mass defect of the reaction (mass of e$^-$ = 5.486$\times10^{-4}$ u)

\begin{align*} \Delta\text{m} &= \underbrace{\text{m}_\gamma}_{\text{product-side}}-\underbrace{2 \text{m}_e}_{\text{reactant-side}}&\text{ positrons and electron have the same mass}\\ &=-2 \text{m}_e&\text{photon does not have mass}\\ &=-2(5.486\times10^{-4} \text{ u}) = -1.0972\times10^{-3}\text{ u}\\ \end{align*}

Convert the mass unit into kg

\begin{align*} -1.0972\times10^{-3}\text{ amu}(1.66056\times10^{-27}\text{ kg/amu}) = -1.822\times10^{-30}\text{ kg} \end{align*}

Calculate the energy of the reaction

\begin{align*} \text{E} &= \Delta\text{mc}^2\\ &=(-1.822\times10^{-30}\text{ kg})(2.9979\times10^8\text{ m/s})^2\\ &=1.637\times10^{-13}\text{ J}\\ \end{align*}

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