KCET · Physics · Electromagnetic Waves
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
- A \( 10^{-10} \mathrm{~cm} \)
- B \( 10^{-12} \mathrm{~cm} \)
- C \( 10^{-14} \mathrm{~cm} \)
- D \( 10^{-16} \mathrm{~cm} \)
Answer & Solution
Correct Answer
(B) \( 10^{-12} \mathrm{~cm} \)
Step-by-step Solution
Detailed explanation
Distance of closest approach is given as
\(d=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(Z_{1} e\right)\left(Z_{2} e\right)}{\text { kinetic energy }} \)
\( \Rightarrow d=\left(9 \times 10^{9}\right) \times \frac{\left(7.9 \times 1.6 \times 10^{-19}\right) \times\left(2 \times 1.6 \times 10^{-19}\right)}{5 \times 10^{6} \times 1.6 \times 10^{-11}} \)
\( =4.55 \times 10^{-15} \mathrm{~m} \Rightarrow 0.45 \times 10^{-12} \mathrm{~cm} \approx 10^{-12} \mathrm{~cm}\)
Therefore, distance of closest approach is of the order of \( 10^{-12} \mathrm{~cm} \)
\(d=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(Z_{1} e\right)\left(Z_{2} e\right)}{\text { kinetic energy }} \)
\( \Rightarrow d=\left(9 \times 10^{9}\right) \times \frac{\left(7.9 \times 1.6 \times 10^{-19}\right) \times\left(2 \times 1.6 \times 10^{-19}\right)}{5 \times 10^{6} \times 1.6 \times 10^{-11}} \)
\( =4.55 \times 10^{-15} \mathrm{~m} \Rightarrow 0.45 \times 10^{-12} \mathrm{~cm} \approx 10^{-12} \mathrm{~cm}\)
Therefore, distance of closest approach is of the order of \( 10^{-12} \mathrm{~cm} \)
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