A moving positron and electron both with kinetic energy 1 MeV annihilate with each other and emits two gamma photons. If the rest mass energy of an electron is 0 . 51 MeV , the wavelength of each photon is ?

Question

A moving positron and electron both with kinetic energy 1 MeV annihilate with each other and emits two gamma photons. If the rest mass energy of an electron is 0 . 51 MeV , the wavelength of each photon is ?, 5.1 × 10 − 3 Å, 10.2 × 10 − 3 Å, 8.2 × 10 − 3 Å, 6.2 × 10 − 3 Å

MARKS App · Accepted Answer

Energy released in annihilation = 0.51 + 0.51 = 1.02 MeV Initial energy = 1 + 1 = 2 MeV . Therefore, the energy of the two photons is = 1.02 + 2 = 3.02 MeV . Hence energy of each photon is E = 1.51 MeV . Now, according to Duane-Hunt law, the wavelength of a photon of energy E (in eV) is given by λ = hc E in eV Å = 6.62 × 10 − 34 × 3 × 10 8 1.51 × 10 6 Å = 8.2 × 10 − 3 Å

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