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Q.

Neon-23 undergoes beta decay in the following way :

10Ne2311Na23+1e0+v-

 Atomic masses of   23Ne and   23Na are 22.9945u and 22.9898u respectively. Determine the maximum KE of βparticle 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

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The Q value of reaction is
Q=Δm×c2
Δm=22.994522.9898=0.0047 
ΔQ=0.0047×931.5MeV=4.4MeV 
This released energy is shared between electron (beta particle) and antineutrino.
The maximum KE of βparticle is 4.4MeV.
 

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