A deutron of atomic mass 2·0147 a.m.u. and negligible kinetic energy is absorbed by a Li6 nucleus…

Q: A deutron of atomic mass 2·0147 a.m.u. and negligible kinetic energy is absorbed by a Li6 nucleus of mass 6·0169 a.m.u., the intermediate nucleus disintegrates spontaneously into two α – particles, each of mass 4·0039 a.m.u. The energy transferred to each α particle is

(a) 12·08 MeV

(b) 11.08 MeV

(c) 6·04 MeV

(d) 5·54 MeV

Ans: (b)

Sol: The given nuclear reaction is

3Li6 + 1H2  → 2 (2He4 )

Mass defect, Δm = (6.0169+2.0147)-2×4.0039

=(8.0316-8.0078) amu

Δm = 0.0238 amu

Energy available = 0.0238 × 931 MeV

KE of each α particle =(0.0238×931)/2 MeV

= 11.08 MeV

Author: Rajesh Jha

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