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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.

Similar Questions

The nucleus $_{48}^{115}Cd$ after two successive ${\beta ^ - }$ decays will give:

The nucleus produced due to $\alpha$-decay of the nucleus $_Z{X^A}$ is

$A$ radioactive decay chain starts from $_{93}Np^{237}$ and produces $_{90}Th^{229}$ by successive emissions. The emitted particles can be:

$A$ nucleus decays by ${\beta ^ + }$ emission followed by a gamma emission. If the atomic and mass numbers of the parent nucleus are $Z$ and $A$ respectively, the corresponding numbers for the daughter nucleus are respectively.

In a radioactive decay process, the negatively charged emitted $\beta -$ particles are

Vedclass Test Series

Exam Paper Generator

Online Exam Module