Question

The closest distance of approach of an $\alpha -$ particle travelling with a velocity $v$ towards $\text{Al}(Z=13)$ nucleus is $d$. The closest distance of approach of an $\alpha -$particle travelling with velocity $4v$ towards $\text{Fe}(Z=26)$ nucleus is-

• A. $\frac{d}{16}$
• B. $\frac{d}{8}$
• C. $\frac{d}{4}$
• D. $\frac{d}{2}$

Answer 1

The correct option is B $\frac{d}{8}$

At the distance of closest approach,

${\text{K.E}}_{\alpha }={\text{Electrostatic P.E}}_{\alpha }$

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{r}[\because q_1=Ze;q_2=2e]$

$\therefore r\propto \frac{Z}{v^2}$

$\Rightarrow \frac{r_2}{r_1}=\frac{Z_2}{Z_1}\times {\left[\frac{v_1}{v_2}\right]}^2$

Substituting the given data gives,

$\Rightarrow \frac{r_2}{d}=\frac{26}{13}\times {\left[\frac{v}{4v}\right]}^2=2\times \frac{1}{16}[\because r_1=d]$

$\Rightarrow r_2=\frac{d}{8}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.