Question

How much Kinetic energy released and what is the daughter nucleus in the$\mathit{\alpha }$decay of polonium-$\mathbf{210}$ ?

Short answer

Kinetic energy release in alpha decay of $polonium-210$ is$5.41\text{\hspace{0.17em}MeV}$ , and daughter nucleus is role="math" localid="1658408913570" $Lead-206$.

Step-by-step solution

Given data

In alpha decay of $\mathrm{polonium}-210$.

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

Atomic mass of${}_{84}^{210}\text{Po},{\mathrm{m}}_{{\text{Po}}_0}=209.982848\text{u}$.

Atomic mass of${}_{82}^{206}\text{Pb},{\mathrm{m}}_{\text{Pb}}=205.97444\text{u}$.

Atomic mass of ${}_2^4\mathrm{\alpha },{\mathrm{m}}_{\mathrm{\alpha }}=4.002603\text{u}$.

Definition of Kinetic Energy

The kinetic energy released is, $\mathbf{Q}\mathbf{=}\mathbf{(}{\mathbf{m}}_{\mathbf{i}}\mathbf{-}{\mathbf{m}}_{\mathbf{f}}\mathbf{)}{\mathbf{c}}^{\mathbf{2}}$.

Where,${\mathbf{m}}_{\mathbf{i}}$is the mass of parents’ nucleus,${\mathbf{m}}_{\mathbf{f}}$is the mass of the daughter nucleus, and$\mathbf{c}$is the speed of light in vacuum.

Determine the Kinetic Energy released

Kinetic energy release,$\mathrm{Q}=({\mathrm{m}}_{\mathrm{i}}-{\mathrm{m}}_{\mathrm{f}}){\mathrm{c}}^2$ .

The conversion factor may be used to express ${\mathrm{c}}^2\to \text{MeV}/\text{u}$.

$\begin{array}{c}\mathrm{Q}=({\mathrm{m}}_{\text{Po}}-{\mathrm{m}}_{\text{Pb}}-{\mathrm{m}}_{\mathrm{\alpha }})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ \mathrm{Q}=(209.982848\text{u}-205.97444\text{u}-4.002603\text{u})\left(931.5\frac{\text{McV}}{\text{u}}\right)\\ \mathrm{Q}=5.41\text{MeV}\end{array}$

Kinetic energy release in alpha decay of $\mathrm{polonium}-210$ is $5.41\text{\hspace{0.17em}MeV}$, and daughter nucleus is $\mathrm{Lead}-206$.