Question

An alpha particle of energy $\frac{1}{2}m{v}^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closet approach for the alpha nucleus will be proportional to

• A. $\frac{1}{v^4}$
• B. $\frac{1}{Z^3}$
• C. $v^2$
• D. $\frac{1}{m}$

Answer 1

The correct option is D $\frac{1}{m}$

Charge present on the alpha particle is $q=2e$

Let the closet distance of approach is $r_0$.

At the distance of the closest approach, the electrostatic potential energy between the alpha particle and the target nucleus is given by,

$U=\frac{1}{4\pi {\epsilon }_0}\times \frac{2e\times Ze}{r_0}$

From the conservative of energy, we have,

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\times \frac{2e\times Ze}{r_0}$

$\Rightarrow r_0=\frac{1}{4\pi {\epsilon }_0}\times \frac{2Z{e}^2}{m{v}^2}$

$\Rightarrow r_0\propto \frac{1}{m}$

Hence, option $\left(D\right)$ is correct.