Calculate the amount of energy released during the $\alpha$-decay $^{238}_{92}\mathrm{U} \rightarrow\ ^{234}_{90}\mathrm{Th} +\ ^{4}_{2}\mathrm{He}$. Given: atomic mass of $^{238}_{92}\mathrm{U} = 238.05079\ u$, $^{234}_{90}\mathrm{Th} = 234.04363\ u$, $^{4}_{2}\mathrm{He} = 4.00260\ u$; $1\ u = 931.5\ \mathrm{MeV}/c^2$. Is this decay spontaneous? Give reason.

$Q \approx 4.25\ \text{MeV}$; spontaneous because $Q>0$.

CBSE Class 12 · 2007 · Delhi · Modern PhysicsShort Answer

Calculate the amount of energy released during the

α

-decay

_{92}^{238} U \to_{90}^{234} Th +_{2}^{4} He

. Given: atomic mass of

_{92}^{238} U = 238.05079 u

_{90}^{234} Th = 234.04363 u

_{2}^{4} He = 4.00260 u

;

1 u = 931.5 MeV / c^{2}

. Is this decay spontaneous? Give reason.

Answer: $Q \approx 4.25 MeV$ ; spontaneous because $Q > 0$ .

Mass defect:

Δ m = 238.05079 - 234.04363 - 4.00260 = 0.00456 u .

Energy released:

Q = 0.00456 \times 931.5 \approx 4.25 MeV .

Since

Q > 0

(energy is released), the decay is spontaneous.

Q \approx 4.25 MeV; spontaneous (Q > 0)

CBSE Class 12 Board · 2007 · Delhi · Subjective