Sum

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 1.097×10^{7} 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 `

Concept: The Line Spectra of the Hydrogen Atom

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