Find out the wavelength of the second Balmer line of `He^(+)` ion .

Explanation

We Know that `1/lambda = z^2 xx R(1)/(n_(1)^(2))-1/(n_(2)^(2))`
For the second Balmer line of `He^+` ion, `n_1=2,n_2=4`
and z=2,Rydberg's constant R= 109678 `cm^(-1)`
`therefore 1/lambda =2^2 xx 109678(1/2^2=1/4^2)`
of `lambda = 12.16 xx 10^(-6)cm = =1216 Å [ because 1 Å = 10^(-8) cm ]`