Question

If the velocity with which an $\alpha$- particle should travel towards the nucleus of a copper atom so as to arrive at a distance 10${}^{-3}$ m from the nucleus of the copper atom is $x\times {10}^{-6}$ m s${}^{-1}$, then the value of closest integer to $x$ is:

Answer 1

Potential energy of the $\alpha$-particle at a distance 10${}^{-13}$ m from the nucleus of a coppor atom is V= $\frac{-Z_1{Z}_2{e}^2}{\left(4\pi {\epsilon }_0\right)r}$= $\frac{-\left(29\right)\left(4\right)(1.6\times {10}^{-19}{C}^2{m}^{-1})}{\left(4\right)\left(3.14\right)(8.85\times {10}^{12}{J}^{-1}{C}^2{m}^{-1}({10}^{-13}m)}$ $=-2.67\times {10}^{-13}$ J
Velocity at which $\alpha$-particle should move is v = $\sqrt{\frac{2|v|}{m}}$= $[\frac{\left(2\right)(2.67\times {10}^{-13}J)}{(4.0\times {10}^{-3}kgmo{l}^{-1})/(6.023\times {10}^{23}mo{l}^{-1})}{]}^{\frac{1}{2}}$
$=8.97\times {10}^6$ m s${}^{-1}$