(a) Using Bohr's second postulate of quantisation of orbital angular momentum, show that the circumference of the electron in the nth orbital state in hydrogen atom is a times the de Broglie wavelength associated with it.
(b) The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally move to the ground state ?

Explanation

(a) As per Bohr.s second postulate the angular momentum of orbiting electron is an integer multiple of `(h)/(2pi) ` i.e., `m v r = n(h)/(2pi) `, where n = 1,2,3,……
`rArr" " 2 pi r = (n h)/(m v )`
But `(h)/(m v) = lambda_(e) = de -` Broglie wavelength associated with electron and ` 2 pi r ` = circumference of nth electron orbit. Thus, it is proved that circumeferene of nth electron orbit is n times the de - Broglie wavelenght assoicated with it .
(b) As the electron state , it is in n = 4 orbit. Thus , it is proved that circumference of nth electron orbits is n times the de - Broglie wavelenght associated with it .
(b) As the electron is in the third excited state , it is in n = 4 orbit. Hence , maximum number of spectral lines which be emitted before it comes to ground state ` = (1)/(2) n(n-1) = (1)/(2) xx 4xx 3 = 6` .