Physics - Modern Physics Question with Solution | TestHub

In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5\times 10^{-14}\mathrm{m}$. If target nucleus has atomic number 80 , then maximum velocity of $\alpha$ - particle is _______ $\times 10^5\mathrm{m}/\mathrm{s}$ approximately. $(\frac{1}{4\pi {ϵ}_0}=9\times 10^9$ SI unit, mass of $\alpha$ particle $=6.72\times 10^{-27}\mathrm{kg})$