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प्रश्न
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?
उत्तर
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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