Question

what transition in hydrogen spectrum would have the same wavelength as the balmer transition n=4 to n=2 of he+ spectrum

Answer 1

The wave number for Balmer transition of n =4 to n=2 of He⁺ spectrum would be:-

$\overline V = RZ^{2} \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$

$\overline V = \dfrac{1}{\lambda} = R(2)^{2} \left( \dfrac{1}{4} - \dfrac{1}{6} \right)$

$\overline V = \dfrac{1}{\lambda} = 4R \left( \dfrac{4-1}{16} \right)$

$\overline V = \dfrac{1}{\lambda} = \dfrac{3R}{4}$

$\lambda = \dfrac{4}{3R}$

Acc. to question, the transition for Hydrogen will have the same wavelength:-

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

$\left[ \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right] = \dfrac{3}{4}$

This equation can only be true when $n_1 = 1$ and $n_2 = 2$

$\therefore$ The transition of electron from n = 2 to n = 1 in Hydrogen would have the same wavelength as transition of electron from n = 4 to n = 1 in $He^{+}$

Answer 2

Answer:

transition n2=2 to n1=1 will give spectrum of the same wavelength as that of Balmer transition, n2=4 to n1=2 in He+