Question

Find the frequency of revolution of an electron in Bohr's $2$nd orbit; if the radius and speed of electron in that orbit is $2.14\times {10}^{-10}$m and $1.09\times {10}^{6}m/s$ respectively. $[\pi =3.142]$.

Answer 1

Radius of $2$nd orbit $r=2.14\times {10}^{-10}$ m

Speed of electron in $2$nd orbit $v=1.09\times {10}^{6}m/s$

Frequency of revolution of an electron in that orbit $\nu =\frac{v}{2\pi r}$

$\therefore$ $\nu =\frac{1.09\times {10}^6}{2\left(3.142\right)(2.14\times {10}^{-10})}$

$⟹$ $\nu =8.1\times {10}^{14}$ Hz