For a single electron atom or ion the wave number of radiation emitted during the transition of electron from a higher energy state `(n = n_(2))` to a lower energy state `(n=n_(1))` is given by the expression:
`bar(v) = (1)/(lambda) = R_(H).z^(2) ((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` ...(1)
where `R_(H) = 2(pi^(2)mk^(2)e^(4))/(h^(3)c) =` Rydberg constant for H-atom
Where the terms have their usual meanings. Considering the nuclear motion, the most accurate expression would have been to replace mass of electron (m) by the reduced mass `(mu)` in the above expression, defined as
`mu = (m'.m)/(m'+m)` where `m'=` mass of nucleus
For Lyman series: `n_(1) =1` (fixed for all the lines)
while `n_(2) = 2,3,4`... for successive lines i.e. `1^(st), 2^(nd),3^(rd)`... lines, respectively. For Balmer series: `n_(1) = 2` (fixed for all the lines) while `n_(2) = 3,4,5`... for successive lines.
The ratio of the wave numbers for the highest energy transition of `e^(-)` in Lyman and Balmer series of H-atom is:

Options

A. `4:1`
B. `6:1`
C. `9:1`
D. `3:1`

Answer

A

Explanation

`bar(v) = R_(H)z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`