Question

Find the wavelength of the radiation emitted by hydrogen in the transitions (a) n = 3 to n= 2, (b) n = 5 to n = 4 and (c) n = 10 to n = 9.

Answer 1

From Balmer empirical formula, the wavelength `(lamda)` of the radiation is given by

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

Here, R = Rydberg constant = `1.097 xx 10^7 m^-1`

           n₁ = Quantum number of final state
           n₂ = Quantum number of initial state

(a)

For transition from n = 3 to n = 2:

Here,

n₁ = 2

n₂ = 3

`1/lamda = 1.09737xx10^7xx (1/4 - 1/9)`

`rArr lamda = 36/(5xx1.0973xx10^7`

`= 6.56 xx 10^-7 = 656  nm`

(b)

For transition from n = 5 to n = 4:

Here,

n₁ = 4

n₂ = 5

`1/lamda = 1.09737 xx 10^-7 (1/16 - 1/25)`

`rArr = 400/(1.09737xx10^7xx9)`

= 4050 nm

(c)

For transition from n = 10 to n = 9:

Here,

n₁ = 9

n₂ = 10

`1/lamda = 1.09737 xx 10^7 (1/81 - 1/100)`

`lamda = (81xx100)/(19xx1.09737xx10^7)`

= 38849  nm