99 Percentile Qs Bank for JEE MainPhysicsNuclear Physics
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 ?
Options
- A5.1 × 10 − 3 Å
- B10.2 × 10 − 3 Å
- C8.2 × 10 − 3 Å
- D6.2 × 10 − 3 Å
Correct answer
C. 8.2 × 10 − 3 Å
Step-by-step solution
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