ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 13. Nuclei

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 \(?\)

  1. A \(n+p \rightarrow d+\gamma\)
  2. B \(e ^{+}+ e ^{-} \rightarrow \gamma\)
  3. C \(n + n \rightarrow\) deuterium atom (electron bound to the nucleus)
  4. D \(p \rightarrow n+e^{+}+\bar{v}\)
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.
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app