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.
If proton in H-nucleus be replaced by positron having the same mass as that of electron but same charge as that of proton, then considering the nuclear motion, the wavenumber of the lowest energy transition of `He^(+)` ion in Lyman series will be equal to

Options

A. `2R_(H)`
B. `3R_(H)`
C. `4R_(H)`
D. `R_(H)`

Answer

B

Explanation

`bar(v) = R_(H)z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] = 3R_(H)`