A nucleus of mass 20 u emits a γ -photon of 6 M e V . If the emission assume to occur when nucleus is free and rest, then the nucleus will have kinetic energy nearest to Take , 1 u = 1.6 × 1 0 - 27 k g .

Step-by-step solution

Due to momentum of photon, nucleus possesses recoil momentum, ∵ E = m c 2 = m c c p = m c = p c ∴ Momentum of γ -photon p 1 = E c = p ......(i) According to the law of conservation of linear momentum p i = p f O = p 1 + p 2 Where, p 2 is linear momentum of nucleus ⇒ p 2 = - p 1 = E c [in magnitude] .....(ii) ∵ Kinetic energy = 1 2 m v 2 [From equation (i)] = 1 2 m v 2 m ∴ p = m v = 1 2 E c 2 m [from equation (ii)] ∴ K E = 1 2 6 × 1 0 6 × 1.6 × 10 - 19 / 3 × 10 8 2 20 × 1.6 × 1 0 - 27 = 0.16 × 1 0 - 15 J = 1.6 × 1 0