Electrostatics 2 Question 16
17. An alpha particle of energy $5 \mathrm{MeV}$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of
(a) $1 \AA$
(b) $10^{-10} \mathrm{~cm}$
(c) $10^{-12} \mathrm{~cm}$
(d) $10^{-15} \mathrm{~cm}$
$(1981,2 \mathrm{M})$
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Answer:
Correct Answer: 17. (c)
Solution:
- From conservation of mechanical energy decrease in kinetic energy = increase in potential energy
$ \begin{aligned} & \text { or } \frac{1}{4 \pi \varepsilon_{0}} \frac{(Z e)(2 e)}{r_{\min }}=5 \mathrm{MeV} \\ & =5 \times 1.6 \times 10^{-13} \mathrm{~J} \\ & \therefore \quad r_{\min }=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 Z e^{2}}{5 \times 1.6 \times 10^{-13}} \\ & =\frac{\left(9 \times 10^{9}\right)(2)(92)\left(1.6 \times 10^{-19}\right)^{2}}{5 \times 1.6 \times 10^{-13}} \\ & \begin{array}{l} (+Z e) \quad(Z=92) \\ \end{array} \end{aligned} $
$=5.3 \times 10^{-14} \mathrm{~m}$
$=5.3 \times 10^{-12} \mathrm{~cm}$
i.e. $r_{\min }$ is of the order of $10^{-12} \mathrm{~cm}$.
$\therefore$ Correct option is (c).
