Statement-1: A nucleus having energy E_{1} de | vedclass

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Question

(C) In $\beta^{-}$ decay,a neutron transforms into a proton,an electron,and an antineutrino $(n ightarrow p + e^{-} + \bar{
u}_{e})$. Statement-$1$

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is a correct explanation for Statement-$1$.

Vedclass · Answer

(C) In $\beta^{-}$ decay,a neutron transforms into a proton,an electron,and an antineutrino $(n ightarrow p + e^{-} + \bar{
u}_{e})$. Statement-$1$ is true because the energy released $(Q = E_{1} - E_{2})$ is shared between the electron and the antineutrino. Since the antineutrino carries away a variable amount of energy,the electron's energy spectrum is continuous,with the maximum (end-point) energy equal to $E_{1} - E_{2}$. Statement-$2$ is also true. If only two particles were involved,the conservation of energy and momentum would require the electron to have a fixed kinetic energy (a discrete spectrum). The observation of a continuous spectrum confirms the emission of a third,invisible particle (the antineutrino) to satisfy conservation laws. Thus,Statement-$2$ correctly explains why the energy spectrum in Statement-$1$ is continuous.

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