Question

An alpha nucleus of energy $\frac{1}{2}mv^2$ 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
$\frac{1}{2}mv^{2}$
$=\frac{1}{4\pi\varepsilon_{0}}$
$\frac{(Ze)(2e)}{r_{0}}$
where Ze = charge o f target nucleus
2e =charge o f alpha nucleus,
$\frac{1}{2} mv^{2}$ = kinetic energy o f alpha nucleus o f mass m moving with velocity v
or $r_{0}=\frac{2Ze^{2}}{4\pi\varepsilon_{0}\left(\frac{1}{2}mv^{2}\right)}$
or $r_{0} \propto\frac{1}{m}$