CBSE-XI-Physics
47: The Special Theory of Relativity
- #23An electron and a positron moving at small speeds collide and annihilate each other. Find the energy of the resulting gamma photon.Ans : We know,
Mass of electron = Mass of positron = 9.1 × 10-31 kg
Both are oppositely charged and annihilate each other to form a gamma photon of rest mass zero. Thus,
∆m = me + mp = 2 × 9.1 × 10-31 kg
This mass will be converted into energy of the resulting γ photon. Thus,
Eγ= ∆mc2
Eγ = 2 × 9.1 × 10-31 × 9 × 1016 J
`` =\frac{2\times 9.1\times 9\times {10}^{-15}}{1.6\times {10}^{-19}}``
`` =102.37\times {10}^{4}\,\mathrm{\,eV\,}``
`` =1.02\times {10}^{5}\,\mathrm{\,eV\,}``
`` =1.02\,\mathrm{\,MeV\,}``
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