Question

An $\alpha$-particle of energy $5MeV$ is scattered through ${180}^0$ by a fixed uranium nucleus. The closest distance is in the order of

• A. $1$$A^0$
• B. ${10}^{-10}$ cm
• C. ${10}^{-12}$ cm
• D. ${10}^{-16}$ cm

Answer 1

Answer: (C)

${10}^{-12}$ cm

The distance of closest approach is
$d=\frac{Z{e}^2}{4\pi {ϵ}_0{E}_k}$
Here, Z = 92,
$E_k=5MeV=5\times {10}^6\times 1.6\times {10}^{-19}J$
Hence,
$d=\frac{9\times {10}^9\times 2\times 92(1.6\times {10}^{-19}{)}^2}{5\times {10}^6\times 1.6\times {10}^{-19}}$

$d=\frac{2649.6\times {10}^{-29}}{8\times {10}^{-13}}$

$d=529.9\times {10}^{-16}=0.5299\times {10}^{-12}$
$d\approx {10}^{-12}cm$