In an alpha particle scattering experiment, the distance of closest approach for the $\alpha$ particle is $4.5\times10^{-14}\ \text{m}$. If the target nucleus has atomic number 80, then the maximum velocity of the $\alpha$-particle is ______ $\times10^5\ \text{m/s}$ approximately. $\left(\dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{SI unit}, \text{mass of}\ \alpha\ \text{particle} = 6.72\times10^{-27}\ \text{kg}\right)$
Numerical value type. Enter your answer.
Answer: 156
At closest approach (head-on), all the kinetic energy becomes potential energy:
$$\frac12mv^2 = \frac{k(2e)(Ze)}{r_0}$$
$$v^2 = \frac{2\times9\times10^9\times2\times80\times(1.6\times10^{-19})^2}{6.72\times10^{-27}\times4.5\times10^{-14}} = \frac{7.37\times10^{-26}}{3.02\times10^{-40}} \approx 2.44\times10^{14}$$
$$v \approx 1.56\times10^7\ \text{m s}^{-1} = 156\times10^5\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics