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 electron

• A. From $n_1=6$ to $n_2=4$
• B. From $n_1=5$ to $n_2=3$
• C. From $n_1=4$ to $n_2=2$
• D. From $n_1=3$ to $n_2=2$

Answer 1

The correct option is A From $n_1=6$ to $n_2=4$

Same wavelength for $He(Z=2)$ and $H(Z=1)$ atoms
First line of Balmer series
$(n=3)\to (n=2)$
${\lambda }_H={\lambda }_{H}e$
$1^2(\frac{1}{2^2}-\frac{1}{3^2})=2^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$(\frac{1}{n_1^2}-\frac{1}{n_2^2})=\frac{5}{36}$

Positive Integral solutions possible are: $n_1=6,n_2=4$