Given the masses of various atomic particles m p =1.0072 u , m n =1.0087 u m e =0.000548 u , m overline v =0, m d =2.0141 u where p ≡ proton, n ≡ neutron, e ≡ electron, overline v ≡ antineutrino and d ≡ deuteron. Which of the following process is allowed by momentum and energy conservation ?

Question

Given the masses of various atomic particles m p =1.0072 u , m n =1.0087 u m e =0.000548 u , m overline v =0, m d =2.0141 u where p ≡ proton, n ≡ neutron, e ≡ electron, overline v ≡ antineutrino and d ≡ deuteron. Which of the following process is allowed by momentum and energy conservation ? (A) n+p → d+&gamma; (B) e ++ e -→ &gamma; (C) n + n → deuterium atom (electron bound to the nucleus) (D) p→ n + e ++ overline v

Tardigrade Admin · Accepted Answer

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

$n + p \to d + γ$

$e^{+} + e^{-} \to γ$

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

$p \to n + e^{+} + \overline{v}$

Only in case-I, $M_{L H S} > M_{R H S}$ i.e.
total mass on reactant side is greater then that on the product side. Hence it will only be allowed

Q. Given the masses of various atomic particles mp​=1.0072u , mn​=1.0087u me​=0.000548u , mv​=0, md​=2.0141u where p≡ proton, n≡ neutron, e≡ electron, v≡ antineutrino and d≡ deuteron. Which of the following process is allowed by momentum and energy conservation ?

n+p→d+γ

e++e−→γ

n+n→ deuterium atom (electron bound to the nucleus)

p→n+e++v

Only in case-I, MLHS​>MRHS​ i.e. total mass on reactant side is greater then that on the product side. Hence it will only be allowed

$n + p \to d + γ$

$e^{+} + e^{-} \to γ$

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

$p \to n + e^{+} + \overline{v}$

Only in case-I, $M_{L H S} > M_{R H S}$ i.e.
total mass on reactant side is greater then that on the product side. Hence it will only be allowed