CBSE Class 12 · 2004 · Modern PhysicsShort Answer

When a deuteron of mass

2.0141 u

and negligible kinetic energy is absorbed by a lithium (

_{3}^{6} Li

) nucleus of mass

6.0155 u

, the compound nucleus disintegrates spontaneously into two alpha particles, each of mass

4.0026 u

. Calculate the energy in joules carried by each alpha particle. (

1 u = 1.66 \times 1 0^{- 27} kg

Answer: $\approx 1.82 \times 1 0^{- 12} J$ ( $\approx 11.4 MeV$ ) per $α$ .

Solution

Reaction:

_{1}^{2} H +_{3}^{6} Li \to 2_{2}^{4} He

.
Mass defect:

Δ m = (2.0141 + 6.0155) - 2 (4.0026) = 8.0296 - 8.0052 = 0.0244 u .

Energy released

= Δ m c^{2} = 0.0244 \times 1.66 \times 1 0^{- 27} \times (3 \times 1 0^{8})^{2} \approx 3.64 \times 1 0^{- 12} J

(total

\approx 22.7 MeV

).
As the deuteron's kinetic energy is negligible and two identical

α

-particles share the energy equally,

E_{α} = \frac{3.64 \times 1 0 ^{- 12}}{2} \approx 1.82 \times 1 0^{- 12} J (\approx 11.4 MeV) .

E_{α} \approx 1.82 \times 1 0^{- 12} J each

CBSE Class 12 Board · 2004 · Subjective