Question

With what velocity (in $\times {10}^{6}m{s}^{-1}$) should an $\alpha$- particle travel towards the nucleus of a copper atom so as to arrive at a distance 10${}^{-13}$ m from the nucleus of the copper atom?

Round-off to the nearest integer.

Answer 1

Potential energy of the $\alpha$-particle at a distance 10${}^{-13}$ m from the nucleus of a copper 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${}^{-1}$