The ratio of angular frequency (ω ) of an electron to the total...

The total energy of an electron in an orbit is given by $E = - \dfrac{1}{2} m v^2$.

The angular frequency is $\omega = \dfrac{v}{r}$, which gives $v = \omega r$.

Substituting $v$ in the energy expression:

$E = - \dfrac{1}{2} m \omega^2 r^2$

The ratio of angular frequency to total energy is:

$\dfrac{\omega}{E} = \dfrac{\omega}{- \dfrac{1}{2} m \omega^2 r^2} = \dfrac{-2}{m \omega r^2}$

From Bohr's quantization rule for angular momentum:

$m v r = \dfrac{nh}{2\pi}$

$m \omega r^2 = \dfrac{nh}{2\pi}$

For the first orbit of the hydrogen atom, $n = 1$:

$m \omega r^2 = \dfrac{h}{2\pi}$

Substituting this value in the ratio expression:

$\dfrac{\omega}{E} = \dfrac{-2}{\dfrac{h}{2\pi}} = -\dfrac{4\pi}{h}$

Answer: $- 4\pi / h$