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 ?

(B) e ^{+}+ e ^{-} \rightarrow \gamma

(C) n + n \rightarrow deuterium atom (electron bound to the nucleus)

(D) p \rightarrow n+e^{+}+\bar{v}

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JEE Mains · Physics · STD 12 - Nuclei

JEE Mains 2020
Medium
MCQ

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 \(p \equiv\) proton, \(n \equiv\) neutron, \(e \equiv\) electron, \(\overline{ v } \equiv\) antineutrino and \(d \equiv\) deuteron. Which of the following process is allowed by momentum and energy conservation \(?\)

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(A) \(n+p \rightarrow d+\gamma\)

Step-by-step Solution

Detailed explanation

Only in \(case-I,\) \(M _{ LHS }> M _{ RHS }\) i.e.

total mass on reactant side is greater then that on the product side. Hence it will only be allowed.

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