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Q.

In the Geiger-Marsden scattering experiment, find the distance of closest approach to the nucleus of a 7.7Mev $α$ -particle before it comes momentaily to rest and reverses its direction.
(‘Z’ for gold nucleus = 79 )

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An Intiative by Sri Chaitanya

a

10 fm

b

20 fm

c

30 fm

d

40 fm

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {mv}^{2} = \frac{1}{4 {πϵ}_{0}} \frac{q_{1} q_{2}}{r_{0}} \\ 7 .7 \times 1 .6 \times 10^{- 13} \\ = \frac{9 \times 10^{9} \times 79 \times 1 .6 \times 10^{- 19} \times 2 \times 1 .6 \times 10^{- 19}}{r_{0}} \\ r_{0} = 29.55 fm \approx 30 fm \end{array}$

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