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Question

Originally the radioactive beta decay was thought as a decay of a nucleus with the emission of electrons only (Case I). However, in addition to the electron, another (nearly) massless and electrically neutral particle is also emitted (Case II). Based on the figure, which of the following is correct :
1688486_07a28cb0113f43ccbb4f378cb02cf7ee.png

A
(a) in both case I and II
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B
(a) in case I and (b) in case II
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C
(a) in case II and (b) in case I
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D
(b) in both cases I and II
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Solution

The correct option is B (a) in case I and (b) in case II
In case-I, no neutrino, or antineutrino is coming out so energy of E-particle will be same for all the decays. In case-II, since neutrino or anti-neutrino is also coming out so energy of E-particle will become variable.

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