What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
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)`
n1 = 2
n2 = 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
n1 = 1and n2 = 2.
∴ The transition for n2 = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He+ spectrum.