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For one-electron species, the wave numbe...

For one-electron species, the wave number of radiation emitted during the transition of electron from a higher energy state `(n_(2))` to a lower energy state `(n_(1))` is given by: `bar v =(1)/(lamda)=R_(H) xx Z^() ((1)/(n_(1)^(2)) (1)/(n_(2)^(2)))` where `R_(H)=(2 pi m_(s) k^(2) c^(4))/(h^(3) c)` is Rydberg constant for hydrogen atom. Now, considering nuclear motion, the accurate measurement would be obtained by replacing mass of electron `(m_(e))` by the reduced mass `(mu)` in the above expression, defined as `mu =(m_(n)xx m_(e))/(m_(n) +m_(e))` where `m_(n)` = mass of nucleus. For Lyman series, `n_(t) =1` (fixed for all the lines) while `n_(2) = 2, 3, 4`.... For Balmer series: `n_(1) = 2` (fixed for all the lines) while `n_(2) = 3,4,5`....
The ratio of the wave numbers for the highest energy transition of electron in Lyman and Balmer series of hydrogen atom is

A

`4:1`

B

`6:1`

C

`9:1`

D

`3:1`

Text Solution

Verified by Experts

The correct Answer is:
A
Lyman series, `n_(1)=1, (1)/(lamda_(1)) = barv_(1) = R_(H).1^(2) [(1)/(1^(2))-(1)/(n_(2)^(2))],` Balmer series, `n_(2) =1, (1)/(lamda_(2)) = bar v = R_(H).1^(2) [(1)/(2^(2))-(1)/(n_(2)^(2))]` For highest energy transition For Lyman series `n_(2)=oo` For Balmer series `n_(2)=oo`
Therefore, `(barv_(1))/(barv_(2)) =(4)/(1)`
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