The speed of an electron in the inner most orbit of the hydrogen (Bohr radius = 52.9 pm; me$=9.11\times {10}^{-31}\mathrm{kg}$) is

Solution: According to Bohr’s atomic model one of the postulates it states that in an atom the angular momentum of an electron in the ${\mathrm{n}}^{\mathrm{th}}$ orbit should be equal to the integral multiple of $\frac{\mathrm{h}}{2\mathrm{\pi }}$. So,$\mathrm{mvr}=\frac{\mathrm{nh}}{2\mathrm{\pi }}.$Then velocity of an electron in ${\mathrm{n}}^{\mathrm{th}} \mathrm{orbit} \mathrm{is} \mathrm{V}=\frac{\mathrm{nh}}{2\mathrm{\pi mr}}\to \mathrm{eqn}\left(1\right)$

Now, substitute n =1and the remaining all constant values in the equation 1.

$\mathrm{V}=\frac{1\times 6.625\times {10}^{-34}}{2\times 3.14\times 9.1\times {10}^{-31}\times 0.529\times {10}^{-10}}$

$\therefore \mathrm{V}=2.19\times {10}^6\mathrm{m}/\mathrm{s}.$