(i) Using Bohr’s second postulate of quantisation of orbital, angular momentum, show that the circumference of the electron in the rath orbital state in H-atom is ra times the de-Broglie wavelength associated with it.

(ii) The electron in H-atom is initially in the . third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?

(i) Bohr’s second postulate states that the electron revolves around the nucleus in certain privileged orbit which satisfy certain quantum condition that angular momentum of an electron is an integral multiple of h / 2$\pi$ , where h is Planck’s constant.

i.e. L = mvr = nh / 2 $\pi$

where, m = mass of electron, v = speed of electron and r = radius of orbit of electron.

2 $\pi$ r = n (h/mv)

Circumference of electron in nth orbit = n x de - broglie wavelength associated with electron.

(ii) Given, the electron in H-atom is initially in third excited state.

n = 4

And the total number of spectral lines of an atom that can exist is given by the relation = n(n-1)/ 2

Here,n=4

So, number of spectral lines = 4 (4-1)/ 2 = 4 x 3/2 = 6

Hence, when a H-atom moves from third excited state to ground state, it emits six spectral lines.