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Q.

Neon-23 undergoes beta decay in the following way : 10Ne23→ 11Na23+ −1e0+v- Atomic masses of   23Ne and   23Na are 22.9945u and 22.9898u respectively. Determine the maximum KE of β−particle in MeV  1u=931.5MeV/c2

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a

4.4MeV

b

1.2MeV

c

0

d

8.8 MeV

answer is A.

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Detailed Solution

The Q value of reaction isQ=Δm×c2Δm=22.9945−22.9898=0.0047 ΔQ=0.0047×931.5MeV=4.4MeV This released energy is shared between electron (beta particle) and antineutrino.The maximum KE of β−particle is 4.4MeV.
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