Q.

A free nucleus of mass 24 amu emits a gamma photon (when initially at rest). The energy of the photon is 7 MeV. The recoil energy of the nucleus in keV is

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a

2.2

b

1.1

c

3.1

d

22

answer is B.

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Detailed Solution

2mk=Ec⇒k=E22mc2=(7MeV)22×24×1.67×10−27×(3×108)2=1.1KeV
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