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Question
Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).
Solution
Given:
Possible transitions:
From n1 = 1 to n2 = 3
n1 = 2 to n2 = 4
(a) Here, n1 = 1 and n2 = 3
Energy, `E = 13.6 (1/n_1^2- 1/n_2^2)`
`E = 13.6 (1/1 - 1/9)`
`= 13.6xx8/9 ....... (i)`
Energy (E) is also given by
`E = (hc)/lamda`
Here, h = Planck constant
c = Speed of the light
λ = Wavelength of the radiation
`therefore E = (6.63xx10^-34xx3xx10^2)/lamda` ........(i)
`"Equanting equation (1) and (2) , we have"`
`lamda = (6.63xx10^-34xx3xx10^8xx9)/(13.6xx8)`
= 0.027 × 10-7 = 103 nm
(b) Visible radiation comes in Balmer series.
As 'n' changes by 2, we consider n = 2 to n = 4.
`"Energy" , E_1 = 13.6 (1/n_1^2 - 1/n_2^2)`
= `13.6 xx (1/4 - 1/16)`
=2.55 eV
If `lamda_1` is the wavelength of the radiation, when transition takes place between quantum number n = 2 to n = 4, then
`255 = 1242/lamda_1`
or λ1 = 487 nm
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