An -particle after passing through a potential difference of V vo… - Vidyadip

MCQ

An $\alpha$-particle after passing through a potential difference of $V$ volts collides with a nucleus. If the atomic number of the nucleus is $Z$, then the distance of closest approach of $\alpha$-particle to the nucleus will be

✓
$14.4 \frac{Z}{V} Å$
B
$14.4 \frac{ Z }{ V } m$
C
$14.4 \frac{ Z }{ V } cm$
D
$1.44 \times 10^{-15} m$

✓

Answer

Correct option: A.

$14.4 \frac{Z}{V} Å$

(A) $14.4 \frac{Z}{V} Å$
For closest approach, Initial kinetic energy of $\alpha$-particle = Potential energy of $\alpha$-particle at closest approach
$
\begin{array}{l}
(2 e) V=\frac{1}{4 \pi \varepsilon_0} \frac{(2 e)(Ze)}{d} \\
d=\frac{1}{4 \pi \varepsilon_0} \frac{Ze}{V}=\left(9 \times 10^9\right) \frac{Z}{V}\left(1.6 \times 10^{-19}\right)=14.4 \frac{Z}{V} Å
\end{array}
$

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