Question

Consider the beta decay

${}^{198}\text{Au}\to {}^{198}{\text{Hg}}^{\ast }+{\beta }^{-}+\overline{\nu }$

Where ${}^{198}{\text{Hg}}^{\ast }$ represents a mercury nucleus in an excited state at energy $1.088\text{MeV}$ above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic mass of ${}^{198}\text{Au}$ is $197.968233u$ and that of ${}^{198}\text{Hg}$ is $197.966760u$. $c^2=931\frac{\text{MeV}}{u}$

• A. $0.28\text{MeV}$
• B. $5.86\text{MeV}$
• C. $9.25\text{MeV}$
• D. $15.89\text{MeV}$

Answer 1

The correct option is A $0.28\text{MeV}$

If the product nucleus ${}^{198}\text{Hg}$ is formed in its ground state, the kinetic energy available to the electron and the antineutrino are given by,

$O=\Delta m{c}^2$

$Q=\left[m{(}^{198}\text{Au}\right)-m{(}^{198}\text{Hg}\left)\right]c^2$

As ${}^{198}Hg\ast$ has energy $1.088\text{MeV}$ more than ${}^{198}\text{Hg}$ in ground state, the kinetic energy actually available will be

$Q=\left[m{(}^{198}\text{Au}\right)-m{(}^{198}\text{Hg}\left)\right]c^2-1.088\text{MeV}$

$Q=(197.968233u-197.966760u)\left(931\frac{\text{MeV}}{u}\right)-1.088\text{MeV}$

$Q=1.3713\text{MeV}-1.088\text{MeV}$

$Q=0.2833\text{MeV}$

This will be the maximum possible kinetic energy of the electron emitted.

Hence, option $\left(A\right)$ is corect.

Why this Question? ${\beta }^{-}decay:$ ${}_Z{X}^A{\to }_{Z+1}{Y}^A{+}_{-1}{e}^0+Q$ ${}_{-1}{e}^0$ can also be written as ${}_{-1}{\beta }^0$