In cgs units,

velocity of an electron (V) in the n th orbit of hydrogen like atom with atomic no. z is given by , V_n = 2 ze² / (nh)

where, e = magn. of charge of electron = 4.8x10^-10 esu of charge;

n = priciple quantum no. h = planck's constant = 6.626x10 ^-27 ergs. sec

 

For hydrogen atom, z = 1,

and V_n = 2 . 1. (4.8 x 10 ^{- 10})² / (6.626 x 10 ^{- 27}) n       cm/s

            = 2.18  x 10⁸ / n              cm / s

 

By the given conditions of the problem,

V_n / (vel. of light) = 1 / 275

i.e 2.18 x 10⁸ / ( n x 3 x 10¹⁰ ) = 1/ 275

[ since vel. of light = 3 x 10¹⁰ cm / s ]

i.e n = 2    ( approximately)

 

So, the reqd. principle quantum no. of the orbit is 2.

 

Now, the wave number of electromagnetic radiation of wavelength  is given by

 

1 /  

 

Again,

 

1 /  = Rz² [ 1/ n₁² - 1 / n₂² ]

 

Here, z = 1 (as it is hydrogen atom)

         n₁ = 1 (as the electron returns to the ground state)

        n₂ = 2 ( as calculated ),

        R = Rydberg constant = 109737 cm ^{- 1}

 

1 /  = 109737 . [ 1 - 1/4 ] cm ^-1 = 82303 cm ^-1

 

Ans: Principle quantum no. is 2 and wave no. of the radiation when the electron comes back to the ground state is 82303 cm^-1

 

Cheers !!