Question

A positron and an electron can annihilate each other on colliding, producing energy as photons:

10e++10e200γ{_{-1}^{0} e}+{_{+1}^{0} e} \longrightarrow 2_{0}^{0} \gamma

Assuming that both γ\gamma rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

Solution

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Answered 2 years ago
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We are given the reaction:

10e + +10e 2 00γ\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^- = 5.486×104\times10^{-4} u)

Δm=mγproduct-side2mereactant-side positrons and electron have the same mass=2mephoton does not have mass=2(5.486×104 u)=1.0972×103 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

1.0972×103 amu(1.66056×1027 kg/amu)=1.822×1030 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

E=Δmc2=(1.822×1030 kg)(2.9979×108 m/s)2=1.637×1013 J\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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