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Question
Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?
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Solution
As the second photon emitted lies in the Lyman series, the transition will be from the states having quantum numbers n = 2 to n = 1.
Wavelength of radiation `(lamda)` is given by
`1/lamda = R (1/n_1^2 - 1/n_2^2)`
Here, R is the Rydberg constant, having the value of 1.097×107 m-1.
`1/lamda = 1.097 xx 10^7 [1/(1)^2 - 1/(2)^2]`
`1/lamda = 1.097xx10^7 [1 - 1/4]`
`rArr 1/lamda = 1.097 xx 3/4 xx 10^7`
`rArr lamda = 4/(1.097xx3xx10^7)`
=`1.215xx10^-7`
= `121.5 xx 10^-9 = 122 nm `
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