Electrostatics 2 Question 16
17. An alpha particle of energy $5 MeV$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of
(a) $1 \AA$
(b) $10^{-10} cm$
(c) $10^{-12} cm$
(d) $10^{-15} cm$
$(1981,2 M)$
Assertion and Reason
Mark your answer as
(a) If Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I.
(b) If Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.
(c) If Statement I is true; Statement II is false.
(d) If Statement I is false; Statement II is true.
Show Answer
Answer:
Correct Answer: 17. (a, b, c, d)
Solution:
- From conservation of mechanical energy decrease in kinetic energy = increase in potential energy
$$ \begin{aligned} & \text { or } \frac{1}{4 \pi \varepsilon _0} \frac{(Z e)(2 e)}{r _{\min }}=5 MeV \\ & =5 \times 1.6 \times 10^{-13} J \\ & \therefore \quad r _{\min }=\frac{1}{4 \pi \varepsilon _0} \frac{2 Z e^{2}}{5 \times 1.6 \times 10^{-13}} \\ & =\frac{\left(9 \times 10^{9}\right)(2)(92)\left(1.6 \times 10^{-19}\right)^{2}}{5 \times 1.6 \times 10^{-13}} \\ & \begin{array}{l} (+Z e) \quad(Z=92) \\ = \\ =5.3 \times 10^{-14} m \\ =5.3 \times 10^{-12} cm \end{array} \end{aligned} $$
i.e. $r _{\min }$ is of the order of $10^{-12} cm$.
$\therefore$ Correct option is (c).
