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An a-particle of energy \( 5 \mathrm{MeV} \) is scattered through \( 180^{\circ} \) by gold nucleus. The distance of
closest approach is of the order of
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closest approach is of the order of
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Verified Answer
The correct answer is:
\( 10^{-12} \mathrm{~cm} \)
Distance of closest approach is given as
\[
\begin{array}{l}
d=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(Z_{1} e\right)\left(Z_{2} e\right)}{\text { kinetic energy }} \\
\Rightarrow d=\left(9 \times 10^{9}\right) \times \frac{\left(7.9 \times 1.6 \times 10^{-19}\right) \times\left(2 \times 1.6 \times 10^{-19}\right)}{5 \times 10^{6} \times 1.6 \times 10^{-11}} \\
=4.55 \times 10^{-15} \mathrm{~m} \quad \Rightarrow \quad 0.45 \times 10^{-12} \mathrm{~cm} \approx 10^{-12} \mathrm{~cm}
\end{array}
\]
Therefore, distance of closest approach is of the order of \( 10^{-12} \mathrm{~cm} \)
\[
\begin{array}{l}
d=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(Z_{1} e\right)\left(Z_{2} e\right)}{\text { kinetic energy }} \\
\Rightarrow d=\left(9 \times 10^{9}\right) \times \frac{\left(7.9 \times 1.6 \times 10^{-19}\right) \times\left(2 \times 1.6 \times 10^{-19}\right)}{5 \times 10^{6} \times 1.6 \times 10^{-11}} \\
=4.55 \times 10^{-15} \mathrm{~m} \quad \Rightarrow \quad 0.45 \times 10^{-12} \mathrm{~cm} \approx 10^{-12} \mathrm{~cm}
\end{array}
\]
Therefore, distance of closest approach is of the order of \( 10^{-12} \mathrm{~cm} \)
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