Question

$\alpha -$ particles of 6 MeV energy is scattered back from a silver foil. Calculate the maximum volume where the entire charge of the atom is supposed to be accumulated.

[$Z$ for silver $=47$]

• A. $3.6\times {10}^{-28}{m}^3$
• B. $5.97\times {10}^{-42}{m}^3$
• C. $6.55\times {10}^{-71}{m}^3$
• D. $48\times {10}^{-42}{m}^3$

Answer 1

The correct option is D $48\times {10}^{-42}{m}^3$

Given that-

Energy of the $\alpha$ particles = $6Mev=6\times 1.6\times {10}^{-19}$ J

We know that the energy-

$E=\frac{1}{4\pi {ϵ}_0}\frac{\left(Ze\right)\left(2e\right)}{r}$

by substituting all the known terms in the energy equation.

$r=\frac{9\times {10}^9\times 47\times 2\times (1.6\times {10}^{-19}{)}^2}{6\times {10}^6\times 1.6\times {10}^{-19}}=2.25\times {10}^{14}m$

Maximum volume can be accumulated by charge-

$V=\frac{4}{3}\pi r^3=48\times {10}^{-42}{m}^3$

Hence, the correct option is D.