In a fission event of $^{238}_{92}\mathrm{U}$ by fast moving neutrons, no neutrons are emitted and the final products, after the beta decay of the primary fragments, are $^{140}_{58}\mathrm{Ce}$ and $^{99}_{44}\mathrm{Ru}$. Calculate $Q$ for this process. Neglect the masses of the electrons/positrons emitted during the intermediate steps. Given: $m(^{238}_{92}\mathrm{U}) = 238.05079\ u$; $m(^{140}_{58}\mathrm{Ce}) = 139.90543\ u$; $m(^{99}_{44}\mathrm{Ru}) = 98.90594\ u$; $m(^{1}_{0}n) = 1.008665\ u$.

CBSE Class 12 · 2022 · Modern PhysicsShort Answer

In a fission event of

_{92}^{238} U

by fast moving neutrons, no neutrons are emitted and the final products, after the beta decay of the primary fragments, are

_{58}^{140} Ce

and

_{44}^{99} Ru

. Calculate

Q

for this process. Neglect the masses of the electrons/positrons emitted during the intermediate steps. Given:

m (_{92}^{238} U) = 238.05079 u

;

m (_{58}^{140} Ce) = 139.90543 u

;

m (_{44}^{99} Ru) = 98.90594 u

;

m (_{0}^{1} n) = 1.008665 u

Answer: $Q \approx 231 MeV$ .

The overall process is

_{92}^{238} U +_{0}^{1} n \to_{58}^{140} Ce +_{44}^{99} Ru

.
Mass defect:

Δ m = (238.05079 + 1.008665) - (139.90543 + 98.90594) = 239.059455 - 238.811370 = 0.248085 u .

Q = 0.248085 \times 931.5 \approx 231 MeV .

Q \approx 231 MeV

CBSE Class 12 Board · 2022 · Subjective