(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?

# Using Bohr’s second postulate of quantisation of orbital

**prasanna**#1

**prasanna**#2

(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.