Question

What is the velocity of an electron $(m=9.11\times {10}^{-31}kg)$ in the innermost orbit of the hydrogen atom?
(Bohr radius $=0.529\times {10}^{-10}m$).

Answer 1

${\mu }_n=\frac{n^2{h}^2}{4{\pi }^{2}mK{Ze}^2}$ (radius of $n^{th}$ Bohr ortbit=$r_n$

$m=9.1\times {10}^{-31}kge=1.6\times {10}^{-19}h=6.626\times {10}^{-34}K=9\times {10}^{9}J.m/{coulomb}^2$

For $H$ atom $(Z=1)$$\Rightarrow r_1=\frac{h^2}{4{\pi }^{2}mK{e}^2}{r}_1=0.529\times {10}^{-10}m$