In hydrogen spectrum `L_(alpha)` line arises due to transition of electron from the orbit n=3 to the orbit n=2. In the spectrum of singly ionized helium there is a line having the same wavelength as that of the `L_(alpha)` line. This is due to the transition of electron from the orbit:

Options

A. `n = 3 to n=2`
B. `n = 2 to n = 1`
C. `n = 5 to n = 3`
D. `n = 6 to n = 4`

Answer

D

Explanation

`" "H_(alpha) " "He^+`
`(1)^2 xx R(1/(2^2) - 1/(3^2)) = (2^2) xx R (1/(n_1^2 - 1/(n_2^2)))`
On solving ,`n_1 = 4 and n_2 = 6`