JEE Main20206 Sep 2020Evening ShiftPhysicsNuclear PhysicsActual
Given the masses of various atomic particles m P = 1 , 0072   u , m n = 1 , 0087   u , m e = 0 .000548   u , m v ¯ = 0 , m d = 2 .0141   u , where p = proton, n ≡ neutron, e ≡ electron, v ¯ ≡ antineutrino and d ¯ ≡ deuteron. Which of the following process is allowed by momentum and energy conservation :
Options
- An + n → deuterium atom (electron bound to the nucleus)
- Bp → n + e + + v ¯
- Cn + p → d + γ
- De + + e − → γ
Correct answer
C. n + p → d + γ
Step-by-step solution
For n + p → d + γ , The mass defect, ∆ m = m p + m n - m d ⇒ ∆ m = 1 . 0072 + 1 . 0087 - 2 . 0141 ⇒ ∆ m = 0 . 0018 ⇒ ∆ m > 0 ∴ only   n + p → d + γ   is   possible