Question

With what velocity should an $\alpha$-particle travel (expressed in $n$$\times {10}^{6}mse{c}^{-1}$, take the value of $n$ to the nearest integer) towards the nucleus of a copper atom so as to arrive at a distance ${10}^{-13}$ metre from the nucleus of the copper atom?

Answer 1

$\frac{1}{2}m{u}^2=\frac{1}{4\pi {\epsilon }_0}\times \frac{2Z{e}^2}{r}$

or $u^2=\frac{Z{e}^2}{\pi {\epsilon }_0.m.r}$

$u^2=\frac{29\times (1.6\times {10}^{-19}{)}^2}{3.14\times 8.85\times {10}^{-12}\times (4\times 1.672\times {10}^{-27})\times {10}^{-13}}$

$\therefore u=6.3\times {10}^{6}mse{c}^{-1}\simeq 6\times {10}^{6}m{\sec }^{-1}$.