Calculate the impact parameter of a 5MeV particle scattered by 90° when it approaches a gold nucleus.

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2.27 x 10^-14 m

15.27 x 10^-14 m

23.23 x 10^-14 m

77.2 x 10^-14 m

answer is C.

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Detailed Solution

$\begin{array}{l} b = \frac{{Ze}^{2} \cot (\frac{θ}{2})}{4 π \in_{0} [KE]} \\ = \frac{9 \times 10^{9} \times 79 {(1 .6 \times 10^{- 19})}^{2} \cot (90 / 2)}{5 \times 10^{6} \times 1 .6 \times 10^{- 19}} (∵ \cot 45^{0} = 1) \\ = 2 .27 \times 10^{- 14} \end{array}$

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