#### Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition *n* = 4 to *n* = 2 of He^{+} spectrum?

#### Solution

For He^{+ }ion, the wave number (`barv`)associated with the Balmer transition, *n* = 4 to *n *= 2 is given by:

`barv = 1/lambda =RZ^2(1/n_1^2 - 1/n_2^2)`

Where

*n*_{1} = 2

*n*_{2} = 4

Z = atomic number of helium

`barv = 1/lambda = R(2)^2(1/4 - 1/16)`

`= 4R ((4-1)/16)`

`=barv = 1/lambda = (3R)/4`

`=> lambda = 4/(3R)`

According to the question, the desired transition for hydrogen will have the same wavelength as that of He^{+}.

`=> R(1)^2[1/n_1^2 -1/n_2^2] = (3R)/4`

`[1/n_1^2 - 1/n_2^2] = 3/4 .....(1)`

By hit and trail method, the equality given by equation (1) is true only when

*n*_{1} = 1and *n*_{2} = 2.

∴ The transition for *n*_{2} = 2 to *n* = 1 in hydrogen spectrum would have the same wavelength as Balmer transition *n* = 4 to *n *= 2 of He^{+} spectrum.