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 $n_1=2$
$\therefore \frac{1}{6561}=R(1)^2\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 R}{36}$....(i)
For the second spectral line in the Balmer series of singly ionised helium ion $n_2=4$ and $n_1=2 ; Z=2$
$\therefore \frac{1}{\lambda}=R(2)^2\left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3 R}{4}$....(ii)
Dividing equation (i) and equation (ii) we get
$\frac{\lambda}{6561}=\frac{5 R}{36} \times \frac{4}{3 R}=\frac{5}{27} \therefore \lambda=1215\,\mathring{A}$
So, $n _2=4$
$n _1=2$ is verified.