91Ó°ÊÓ

Nuclear radius depends on mass number of nucleus as follows.

$R=r_0{A}^{1/3}$

Here, $r_0$is a constant $\left(r_0=1.2\text{fm}\right)$and $A$is mass number.

Substitute 1.2 fm for $r_0$and 238 for $A$, calculated the radius of${}^{238}\text{U}$ nucleus.

$\begin{array}{l}R=\left(1.2\text{fm}\right){\left(238\right)}^{1/3}\\ =7.43\text{fm}\end{array}$

Convert the unit for nuclear radius from fm to $m$.

$\begin{array}{l}R=7.43\text{fm}\\ =\left(7.43\text{fm}\right)\left(\frac{1.0×{10}^{−15}\text{m}}{1.0\text{fm}}\right)\\ =7.43×{10}^{−15}\text{m}\end{array}$

Diameter of the ${}^{238}\text{U}$nucleus is,

$\begin{array}{l}D=2R\\ =\left(2\right)\left(7.43×{10}^{−15}\text{m}\right)\\ =14.86×{10}^{−15}\text{m}\end{array}$

From the given data, the de Broglie wavelength of the$\mathrm{Î\pm }$ particle is equal to the diameter of the ${}^{238}\text{U}$nucleus.

The de Broglie wavelength of alpha particle is,

$\mathrm{λ}=\frac{h}{p}$

Here, $h$is Plank's constant and $p$is momentum.

Rewrite this equation for momentum,

$p=\frac{h}{\mathrm{λ}}$

A moving $\mathrm{Î\pm }$particle has only kinetic energy, the relation between momentum and energy of the particle is,

$E=\frac{p^2}{2m}$

Substitute $\frac{h}{\mathrm{λ}}$for $p$,

$\begin{array}{l}E=\frac{{\left(\frac{h}{\mathrm{λ}}\right)}^2}{2m}\\ =\frac{h^2}{2{\mathrm{λ}}^{2}m}\end{array}$

Convert atomic mass of the $\mathrm{Î\pm }$particle from u to $\text{kg}$.

$\begin{array}{l}m=4.00260\text{u}\\ =\left(4.00260\text{u}\right)\left(\frac{1.6605×{10}^{−27}\text{kg}}{1.0\text{u}}\right)\\ =6.646×{10}^{−27}\text{kg}\end{array}$

Substitute $14.86×{10}^{−15}\text{m}$for $\mathrm{λ},6.63×{10}^{−34}\text{J}â‹\dots \text{s}$for $h$and $6.646×{10}^{−27}\text{kg}$for $m$in the

equation $E=\frac{h^2}{2{\mathrm{λ}}^{2}m}$.

$\begin{array}{l}E=\frac{{\left(6.63×{10}^{−34}\text{J}â‹\dots \text{s}\right)}^2}{2{\left(14.86×{10}^{−15}\text{m}\right)}^2\left(6.646×{10}^{−27}\text{kg}\right)}\\ =1.49×{10}^{−13}\text{J}\end{array}$

Convert the units for energy from $\text{J}$to $\text{MeV}$.

$\begin{array}{l}E=1.49×{10}^{−13}\text{J}\\ =\left(1.49×{10}^{−13}\text{J}\right)\left(\frac{1.0\text{eV}}{1.6×{10}^{−19}\text{J}}\right)\left(\frac{1.0×{10}^{−6}\text{MeV}}{1.0\text{eV}}\right)\\ =0.93\text{MeV}\end{array}$

Therefore, the energy of the alpha particle is$0.93\text{MeV}$.