Question

What energy (in MeV) alpha particle has a de Broglie wavelength equal to the diameter of a 238 U nucleus?

Short answer

From the given data, the de Broglie wavelength of the particle is equal to the diameter of the ²³⁸U nucleus.

Therefore the energy of the alpha particle is 0.93 MeV

Step-by-step solution

Step 1

Nuclear radius depends on mass number of nucleus as follows.

$R=r_0{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Here, r₀ is a constant (r₀ =1.2fm) and A is mass number

R =(1.2fm)(238)^1/3

=7.43 fm

=7.43x10^-15m

Diameter of the ²³⁸U nucleus is

D=2R

=14.86X10^-15 M

From the given data, the de Broglie wavelength of the $\infty$ particle is equal to the diameter of the ²³⁸U nucleus.

Step 2

The de Broglie wavelength of alpha particle is,

$\lambda =\frac{h}{p}$

Here, h is plank's constant nd p is momentum.

Rewrite this equation for momentum,

$p=\frac{h}{\lambda }$

A moving $\alpha$particle has only kinetic energy the relation between momentum and energy of the particle is,

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{P}}^2}{2\mathrm{m}} \\ \mathrm{Substitute}\frac{\mathrm{h}}{\mathrm{\lambda }}\mathrm{for}\mathrm{p} \\ \mathrm{E}=\frac{{\left({\displaystyle \frac{\mathrm{h}}{\mathrm{\lambda }}}\right)}^2}{2\mathrm{m}} \\ =\frac{{\mathrm{h}}^2}{2{\mathrm{\lambda }}^2\mathrm{m}} \end{gathered}$$

Step 3

Convert atomic mass of the $\alpha$particle from u to kg

m=6.646x10^-27kg

convert the units for energy from j to MeV

E=0.93 MeV

Therefore the energy of the alpha particle is 0.93 MeV