Question

An electron and a positron (an anti-matter electron) meet and completely annihilate, creating two photons of identical energy. What is the energy of each photon created?

Note: the positron has exactly the same mass as the electron.

		about 512 eV
		about 5 eV
		about 0.51 MeV
		about 5.1 MeV

Answer 1

During annihilation of electron and positron energy released will be:

dQ = dm*c^2

dm = mass of electron + mass of positron - mass of photon

dm = 9.1*10^-31 + 9.1*10^-31 - 0

dm = 2*9.1*10^-31 kg

c = speed of light = 3*10^8 m/sec

So,

dQ = dm*c^2

dQ = 2*9.1*10^-31*(3*10^8)^2

dQ = 1.638*10^-13 J

Since 1 eV = 1.6*10^-19 J, So

dQ = 1.638*10^-13/(1.6*10^-19) = 1023750 eV

energy created by each photon = dQ/2

Q_p = dQ/2 = 1023750/2 = 511875 eV

Q_p = 0.51*10^6 eV = 0.51 eV

Correct option is C.

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