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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

  1. An + n → deuterium atom (electron bound to the nucleus)
  2. Bp → n + e + + v ¯
  3. Cn + p → d + γ
  4. 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

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