Question

Question: The radioactive nuclide ${}_{\mathbf{84}}{}^{\mathbf{210}}\mathbf{Po}$decays by alpha emission to a daughter nuclide. The atomic mass of ${}_{\mathbf{84}}{}^{\mathbf{210}}\mathbf{Po}$ is $\mathbf{209}\mathbf{.}\mathbf{9829}\mathbf{}\mathbf{u}$, and that of the daughter is $\mathbf{205}\mathbf{.}\mathbf{9745}\mathbf{}\mathbf{u}$.

1. Identify the daughter, and write the nuclear reaction for the radioactive decay process.
2. Calculate the total energy released per disintegration (in MeV).
3. Calculate the kinetic energy of the emitted alpha particle.

Short answer

1. The daughter nuclide formed from the alpha decay is ${}_{82}{}^{206}\mathrm{Pb}$. The balanced nuclear reaction representing the decay process is:${}_{84}{}^{210}\mathrm{Po}\to {}_{82}{}^{206}\mathrm{Pb}+{}_2{}^4\mathrm{He}$
2. The energy released per disintegration process is $5.40\mathrm{MeV}$.
3. The maximum kinetic energy of the alpha particle is $8.65x{10}^{13}\mathrm{J}$.

Step-by-step solution

Nuclear reaction representing the radioactive decay process

An alpha particle is a doubly ionized helium atom. When an alpha particle is released in a radioactive decay process the mass number of the daughter nuclei is 4 units less than the parent nuclide and the atomic number is 2 units less than the parent nuclide. Therefore the nuclear reaction for the alpha decay by ${}_{84}{}^{210}\mathrm{Pb}$is:

${}_{84}{}^{210}\mathrm{Po}\to {}_{82}{}^{206}\mathrm{Pb}+{}_2{}^4\mathrm{He}$

The daughter nuclide is ${}_{84}{}^{206}\mathrm{Pb}$.

Energy released in the disintegration process

To calculate the energy released per disintegration for the radioactive decay process, we have to estimate the mass difference involved in the decay process.

$$\begin{gathered} ∆m=\mathrm{m}\left({}_{82}{}^{206}\mathrm{Pb}\right)+\mathrm{m}\left({}_2{}^4\mathrm{He}\right)-\mathrm{m}\left({}_{84}{}^{210}\mathrm{Po}\right) \\ =205.97446\mathrm{u}+4.002603\mathrm{u}-209.98287\mathrm{u} \\ =-0.005805\mathrm{u} \end{gathered}$$

The energy equivalent to this mass difference $=-0.005805\mathrm{u}\times 931.5=-5.40\mathrm{MeV}$

The negative sign indicates that the energy is released.

Therefore energy released$=5.40\mathrm{MeV}$

Energy of alpha particle

Since most of the energy released is taken away by the alpha particle, the maximum kinetic energy of the alpha particle will be approximately equal to the energy released by the decay process.

Maximum kinetic energy of alpha particle$=5.40\mathrm{MeV}=5.40\times 1.602\times {10}^{-13}\mathrm{J}=8.65\times {10}^{-13}\mathrm{J}$