Given the masses of various atomic particles m P = 1 , 0072   u , m n = 1 , 0087   u , m e = 0 …

Given the masses of various atomic particles mP=1,0072 u, mn=1,0087 u, me=0.000548 u, mv¯=0, md=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 :
  1. n+n→ deuterium atom (electron bound to the nucleus)
  2. p→n+e++v¯
  3. n+p→d+γ
  4. e++e−→γ

Solution

For n+p→d+γ,

The mass defect, ∆m=mp+mn-md

⇒∆m=1.0072+1.0087-2.0141

⇒∆m=0.0018

⇒∆m>0

∴only n+p→d+γ is possible

Asked in: JEE Main 2020 (06 Sep Shift 2)

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