Q: For electron moving in n^{th} orbit of H-atom the angular velocity is proportional to

(a) n

(b) 1/n

(c) n^{3}

(d) 1/n^{3}

**Click to See Answer : **

Ans: (d)

Sol: $\large T = \frac{2\pi r}{v}$

$ r \propto \frac{n^2}{Z}$ and $ v \propto \frac{Z}{n}$

$ T \propto \frac{n^3}{Z^2}$

$\omega = \frac{2\pi}{T} \propto \frac{1}{n^3}$