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Q.

Distance of closest approach when a 5.0MeV proton approaches a gold nucleus is (Z=79)

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a

47 fermi

b

39 fermi

c

16 fermi

d

23 fermi

answer is B.

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Detailed Solution

$\begin{array}{l} KE = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r} \\ 5 \times 10^{6} \times 1 .6 \times 10^{- 19} = \frac{9 \times 10^{9} \times e \times 79 e}{r} \end{array}$
r = 23fm

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