Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561\mathring{A}$. If the wavelength of the spectral line in the Balmer series of singly-ionized helium atom is $1215\mathring{A}$ when electron jumps from $n_2$ to $n_1$, then $n_2$ and $n_1$ are

• A. 4, 2
• B. 5,3
• C. 6,3
• D. 6,2

Answer 1

Answer: (A)

We know that $\frac{1}{\lambda }=R{Z}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
The wave length of first spectral line in the Balmer series of hydrogen atom is $6561\mathring{A}$. Here $n_2=3$ and $n_1=2$
$\therefore \frac{1}{6561}=R(1{)}^2\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5R}{36}$....(i)
For the second spectral line in the Balmer series of singly ionised helium ion $n_2=4$ and $n_1=2;Z=2$
$\therefore \frac{1}{\lambda }=R(2{)}^2\left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3R}{4}$....(ii)
Dividing equation (i) and equation (ii) we get
$\frac{\lambda }{6561}=\frac{5R}{36}\times \frac{4}{3R}=\frac{5}{27}\therefore \lambda =1215\mathring{A}$
So, $n_2=4$
$n_1=2$ is verified.