Question

What transition in hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of $H{e}^{+}$ Spectrum.

Answer 1

For$H{e}^{+}$,$Z=2$ for Blamer series ($n_1=2and{n}_2=4$)
$\frac{1}{\lambda }=Z^{2}R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\frac{1}{\lambda }=\left(2^2\right)R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{4}$

For$H^{+}$,$Z=1$ for ($n_1=1and{n}_2=2$ )

$\frac{1}{\lambda }=Z^{2}R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\frac{1}{\lambda }=R(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3R}{4}$

Hence, for hydrogen ($n_1=1and{n}_2=2$ ) transition.