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Question
A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?
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Solution
Wavelength of ultraviolet radiation, `lamda = 102.5 "nm" = 102,5 xx 10^-7 m ^-1`
Rydberg's constant, R = 1.097 ×107 m^-1
Since the emitted light lies in ultraviolet range, the lines will lie in lyman series.
Lyman series is obtained when an electron jumps to the ground state (n1 = 1) from any excited state (n2).
Wavelength of light is given by
`1/lamda = R (1/n_1^2 - 1/n_2^2 )`
`Here , R = Rydberg " constant"`
`rArr 1/(102.5 xx 10^-9 )= 1.097 xx10^7 (1/1^2 - 1/n_2^2) `
`rArr 10^9 / 102.5 = 1.097 xx 10^7 (1 - 1/n_2) `
`rArr 10^9 / 102.5 = 1.097 (1 - 1/n_2)`
`rArr l - 1/(n_2^2) = 100/(102.5xx 1.097) `
`rArr 1/n_2^2 = 1 - 100/(102.5xx1.097) `
`rArr n_2 = sqrt(9.041 = 3`
The transition will be from 1 to 3.
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