Question

A group of hydrogen atoms are prepared in n = 4 states. List the wavelength that are emitted as the atoms make transitions and return to n = 2 states.

Answer 1

There will be three wavelengths.
(i) For the transition from (n = 4) to (n = 3) state
(ii) For the transition from (n = 3) to (n = 2) state
(iii) For the transition from (n = 4) to (n = 2) state

Let `(lamda_1)` be the wavelength when the atom makes transition from (n = 4) state to (n = 2) state._​ 
Here,
n₁ = 2
n₂ = 4
Now, the wavelength `(lamda_1)`  will be
`1/lamda_1 = R (1/n_1^2 - 1/n_2^2)`

`R = 1.097 xx 10^7 m^-1`
`1/lamda_1 = 1.097xx 10^7 xx (1/4 - 1/16)`

`rArr 1/lamda_1 = 1.097 xx 10^7 ((4-1)/(16))`

`rArr 1/lamda_1 = (1.097xx10^7xx3)/16`

`rArr lamda_1 = (16xx10^-7)/(3xx1.097)`
= 4.8617 × 10^-7
= 486.1 × 10^-9
= 487 nm

When an atom makes transition from (n = 4) to (n = 3), the wavelength (λ2) is given by
Here again 
`n_1 = 3`
`n_2 = 4`

`1/lamda_2 = 1.097 xx 10^7 (1/9 - 1/16)`

`rArr 1/lamda_2 = 1.097 xx 10^7 ((16 -9)/144)`

`rArr 1/lamda_2 = (1.097xx 10^7 xx 7)/144`

`rArr lamda_2 = 144/(1.097 xx 10^7 xx 7)`

                      = 1875 nm

Similarly, wavelength (λ2) for the transition from (n = 3) to (n = 2) is given by
When the transition is n₁ = 2 to n₂ = 3:
`1/lamda_3 = 1.097 xx 10^7 (1/4 - 1/9)`

`rArr 1/lamda_3 = 1.097 xx 10^7 ((9-4)/36) `

`rArr 1/lamda_3 = (1.097xx10^7)/36`

`rArr lamda_3 = (36xx10^7xx5)/36`

`rArr lamda_3 = (36xx10^7)/((1.097)xx5) = 656  nm`