CBSE Class 12 · 2006 · All India · Modern PhysicsShort Answer

A neutron is absorbed by a

_{3}^{6} Li

nucleus with the subsequent emission of an alpha particle.
(i) Write the corresponding nuclear reaction.
(ii) Calculate the energy released, in MeV. Given: mass

_{3}^{6} Li = 6.015126 u

; mass (neutron)

= 1.0086654 u

; mass (alpha particle)

= 4.0026044 u

; mass (triton)

= 3.0100000 u

. Take

1 u = 931 MeV / c^{2}

Answer: $^{6} Li + n \to^{4} He +^{3} H$ ; $Q \approx 10.42 MeV$ .

Solution

(i) Reaction:

_{3}^{6} Li +_{0}^{1} n \to_{2}^{4} He +_{1}^{3} H .

(ii) Mass defect:

Δ m = (6.015126 + 1.0086654) - (4.0026044 + 3.0100000) = 7.0237914 - 7.0126044 = 0.011187 u .

Q = 0.011187 \times 931 \approx 10.42 MeV .

Q \approx 10.42 MeV

CBSE Class 12 Board · 2006 · All India · Subjective

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