Question

The distance of closest approach of the particle to the nucleus will be

• A. $6.4\times {10}^{-13}$m
• B. $4.3\times {10}^{-13}$m
• C. $2.1\times {10}^{-13}$m
• D. $3.4\times {10}^{-14}$m

Answer 1

The correct option is D $3.4\times {10}^{-14}$m

The charge on an alpha particle is $+2$.
Hence, $q=2e=3.2\times {10}^{-19}C$
This alpha particle is accelerated by a potential of $2\times {10}^{6}V$.
Hence, energy of the particle $=qV=3.2\times {10}^{-19}\times 2\times {10}^6=6.4\times {10}^{-13}J$
At closest approach, $PE=$ total energy.
i.e.
$6.4\times {10}^{-13}=\frac{1}{4\pi {ϵ}_0}\frac{47\times 2e^2}{d}$

Solving this, we get: $d=3.384\times {10}^{-14}m$