JEE Advance - Chemistry (1993 - No. 8)

For a spectral transition

$${1 \over \lambda } = {R_H}{Z^2}\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right)$$

$$\therefore$$ For He⁺ ion, we have

$${1 \over \lambda } = {R_H}\,.\,{(2)^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {3 \over 4}{R_H}$$ ..... (i)

Now for hydrogen atom

$${1 \over \lambda } = {R_H}\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right)$$ ...... (ii)

Equating equations (i) and (ii), we get

$$\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right) = {3 \over 4}$$

Obviously n₁ = 1 and n₂ = 2. Hence, the transition n = 2 to n = 1 in hydrogen atom will have the same wavelength as the transition, n = 4 to n = 2 in He⁺ species. The transition belongs to Lyman series.