In the spectrum of singly ionized helium , the wavelength of a line abserved is almost the same as the first line of Balmer series of hydrogen . It is due to transition of electron

Options

A. `from n _(1) = 6 to n _(2) = 4`
B. `from n _(5) = 6 to n _(2) = 3`
C. `from n _(1) = 4 to n _(2) = 2`
D. `from n _(1) = 3 to n _(2) = 2`

Answer

A

Explanation

For the first line of Balmer series of hydrogen
`(1)/(lambda) = R ((1)/(2^(2)) - (1)/(3^(2))) = (5R)/(36) implies lambda = (36)/(5 R)`
For singly ionized helium `(Z = 2)`,
`(1)/(lambda') = 4 R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
Given `lambda' = lambda`
For `n_(1) = 6 to n_(2) = 4`
`(1)/(lambda') = 4 R ((1)/(4^(2)) - (1)/(6^(2))) = (20R)/(144) = (5R)/(36)`
It corresponds to transition from `n_(1) = 6 to n_(2) = 4`.