A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 e.V (a) Identify the quantum numbers *n* of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.

#### Solution

(a) The binding energy of hydrogen is given by

`E = 13.6/(n^2)eV`

For binding energy of 0.85 eV,

`n_2^2 = 13.6/0.85`

`n_2 = 4`

For binding energy of 10.2 eV,

`n_2^2 = 13.6/10.2 = 16`

n_{2} = 1.15

⇒ n_{1} = 2

The quantum number of the upper and the lower energy state are 4 and 2, respectively.

(b) Wavelength of the emitted radiation (λ) is given by

`1/lamda = R (1/n_1^2 - 1/n_2^2)`

Here,

R = Rydberg constant

*n*_{1} and* **n*_{2} are quantum numbers.

`therefore 1/lamda = 1.097 xx 10^7 (1/4 - 1/16)`

`rArr lamda = (16)/(1.097xx3xx10^7)`

= 4 .8617 × 10^{-7 }

= 487 nm