Question

(a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n 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 moves to the ground state?

Answer 1

(a) According to Bohr’s second postulate of quantization, the electron can revolve round the nucleus only in those circular orbits in which the angular momentum of the electron is integral multiple of where h is Planck’s constant (= 6.62 × 10−34 Js).
So, if m is the mass of electron an v is the velocity of electron in permitted quantized orbit with radius r thanWhere n is the principle quantum number and can take integral values liken = 1, 2, 3…This is the Bohr’s quantization condition.Now, de-Broglie wavelength is given asWhere λ → is wavelength associated with electron.v is the velocity of electron.h − is the velocity of electron.m − mass of electron
Putting value of v from (2) in (1)Now circumference of the electron in the nth orbital state of Hydrogen atom with radius r is 2πr.
(b) In n is the quantum number of the highest energy level involved in the transitions, then the total number of possible spectral lines emitted isThird excited state means fourth energy level i.e. n = 4. Here, electron makes transition from n = 4 to n = 1 so highest n is n = 4
Thus, possible spectral lines6 is the maximum possible number of spectral lines.