Questions

Neon-23 undergoes beta decay in the following way :
$_{10} N e^{23} \to_{11} N a^{23} +_{- 1} e^{0} + \bar{v}$
Atomic masses of $^{23} N e$ and $^{23} N a$ are 22.9945u and 22.9898u respectively. Determine the maximum KE of $β -$ particle in MeV $(1 u = 931.5 M e V / c^{2})$

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4.4MeV

1.2MeV

8.8 MeV

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detailed solution

Correct option is A

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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Match the following two columns.

Column I	Column II
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(d) After emission of two $α$ and two $β$ particles.	(s) mass number will decrease by 4