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 from `n_(1) = 6 to n_(2) = ^(''**'')`. What is the value of `(''**'')`.

Answer

4

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 implies 4 R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = (5R)/(36)`
`implies n_(2) = 4`