Question

The number of revolution per second made by an electron in the first Bohr orbit of the hydrogen atom is $6.5\times {10}^{3k}$. The value $k$ is (Integer only)

Answer 1

The frequency of revolution of an electron in $n^{th}$ orbit in a hydrogen atom is,

${\nu }_n=\frac{v_n}{2\pi r_n}$

For first orbit, $(n=1)$

${\nu }_1=\frac{v_1}{2\pi r_1}.....\left(1\right)$

$\because r_1=0.53\left(\frac{n^2}{Z}\right)\mathring{A}\text{and}{v}_1=2.18\times {10}^6\left(\frac{Z}{n}\right){\text{ms}}^{-1}$

Putting the values of $\left(r_1\right)$, $\left(v_1\right)$ and $Z=1$ in (1) we get,

${\nu }_1=\frac{2.18\times {10}^6}{2\times 3.14\times 0.53\times {10}^{-10}}=6.5\times {10}^{15}{\text{rev s}}^{-1}$

Given, ${\nu }_1=6.5\times {10}^{3k}{\text{rev s}}^{-1}$

$\therefore k=5$