Question

In the spectrum of singly ionized helium, the wavelength of a line observed is almost the same as the first line of Balmer series of hydrogen. It is due to transition of the electron from $n_1=6$ to $n_2=$"$\ast$". What is the value of "$\ast$"?

Answer 1

$\frac{1}{\lambda }=Z^{2}R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
for singly ionized helium, the wavelength of a line observed = first line of Balmer series of hydrogen
$2^{2}R(\frac{1}{n^2}-\frac{1}{6^2})=R(\frac{1}{2^2}-\frac{1}{3^2})$
$n=4$