Question

A proton is moving towards a Tin foil. Its distance of the closest approach is found to be $30\text{fm}$. What will be its distance of the closest approach if it approaches a Lead foil, with the same speed?
Given: Atomic number of Tin$=50$, Atomic number of Lead$=82$

• A. $49.2\text{f m}$
• B. $55.8\text{f m}$
• C. $39.2\text{f m}$
• D. $35.8\text{f m}$

Answer 1

The correct option is A $49.2\text{f m}$

$d=\frac{4KZ{e}^2}{m{v}^2}$

As other parameters (except $Z$) are not changing, we can write,
$\frac{d_{Tin}}{d_{Lead}}=\frac{Z_{Tin}}{Z_{Lead}}$

$\Rightarrow \frac{30}{d_{Lead}}=\frac{50}{82}$

$\therefore d_{Lead}=49.2\text{f m}$

Hence, $\left(A\right)$ is the correct answer.