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Q.

An $α$ -particle of energy 4 MeV is scattered by gold foil (Z=79). Calculate the maximum volume in which positive charge of the atom is likely to be concentrated.

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a

3.3 x 10^–40 m³

b

7.7 x 10^–40 m³

c

5.5 x 10^–40 m³

d

11 x 10^–40 m³

answer is B.

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Detailed Solution

$\begin{array}{l} E = \frac{1}{4 {πϵ}_{0}} \cdot \frac{2 {Ze}^{2}}{r_{0}} \\ 4 \times 10^{6} \times 1 .6 \times 10^{- 19} = 9 \times 10^{9} \times \frac{2 (79) {(1 .6 \times 10^{- 19})}^{2}}{r_{0}} \\ ∴ r_{0} = 5 .7 \times 10^{- 14} m \to (1) \\ \to V = \frac{4}{3} {πr}_{0}^{3} = \frac{4 \times 3 .14}{3} {(5 .7 \times 10^{- 14})}^{3} \\ = 7 .7 \times 10^{- 40} \end{array}$

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