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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.
Solution
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
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