Question

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

• A. $14.4\frac{Z}{V} \, \overset{^\circ }{A}$
• B. $14.4\frac{Z}{V} \, m$
• C. $14.4\frac{Z}{V} \, cm$
• D. all of these

Answer 1

Answer: (A)

K.E. = P.E. = qV
$\Rightarrow 2eV=\frac{K \left(Z e\right) \left(2 e\right)}{d}=qV$
$\therefore d=\frac{9 \times 1 0^{9} \times Z \times e \times 2 e}{2 e V}$
$\therefore d=\frac{9 \times 1 0^{9} \times 1.6 \times 1 0^{- 19} \times Z}{V}$
$d=14.4\times 10^{- 10}\left(\frac{Z}{V}\right)m$