Question

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

• A. $v^2$
• B. $1/m$
• C. $1/v^4$
• D. $1/Ze$

Answer 1

Answer: (B)

At distance o f closest approach $(r_0)$
Kinetic energy = Potential energy
$\frac{1}{2}m{v}^2$
$=\frac{1}{4\pi {\epsilon }_0}$
$\frac{(Ze)(2e)}{r_0}$
where Ze = charge o f target nucleus
2e =charge o f alpha nucleus,
$\frac{1}{2}m{v}^2$ = kinetic energy o f alpha nucleus o f mass m moving with velocity v
or $r_0=\frac{2Z{e}^2}{4\pi {\epsilon }_0\left(\frac{1}{2}m{v}^2\right)}$
or $r_0\propto \frac{1}{m}$