Question

The nuclide ${}_{\mathbf{4}}{}^{\mathbf{10}}\mathbf{Be}$ undergoes spontaneous radioactive decay to${}_{\mathbf{5}}{}^{\mathbf{10}}\mathbf{B}$ with emission of$\mathbf{\beta }$ particle. Calculate the maximum kinetic energy of the$\mathbf{\beta }$ particle.

Short answer

The kinetic energy of the$\beta$ particle is$0.561 \text{MeV}$

Step-by-step solution

β Emission

In a nucleus when the number of neutrons is excess than the number of protons,$\beta$ decay takes place. When a$\beta$ decay takes place, a neutron is converted to a proton and a electron. The high-energy electron emission is known as the$\beta$ particle. The mass number of the daughter nucleus remains same, while the atomic number of the daughter nucleus increases by 1. An antineutrino is emitted along with the particle.

${}_4{}^{\text{8}}\text{Be}\to {}_5{}^{\text{8}}\text{B +}{}_{\text{- 1}}{}^{\text{0}}{\text{e}}^{-}\text{+}\overline{\nu }$

Kinetic energy

To calculate the kinetic energy involved in the$\beta$ decay we have to calculate the mass difference.

$$\begin{gathered} \Delta \text{m}=\text{m}\left({}_5{}^{\text{10}}\text{B}\right)\text{- m}\left({}_4{}^{\text{10}}\text{B}\right) \\ =10.01293 \text{u - 10.013533u} \\ =-0.0006031 \text{u} \end{gathered}$$

$$\begin{gathered} \Delta E=-0.000603\times 931.5 \text{MeV} \\ =-0.561 \text{MeV} \end{gathered}$$

The energy released in the nuclear reaction is$-0.561 \text{MeV}$. This energy is carried out by the electron. Therefore, the kinetic energy of the particle is$0.561 \text{MeV}$.