IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# Bohr's Theory of Hydrogen-like Atoms

## Problems from IIT JEE

Problem (IIT JEE 2011): The wavelength of the first spectral line in the Balmer series of hydrogen atom is ${6561}\;{Angstrom}$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

1. ${1215}\;{Angstrom}$
2. ${1640}\;{Angstrom}$
3. ${2430}\;{Angstrom}$
4. ${4687}\;{Angstrom}$
The wavelength of spectral line when an electron in singly ionized atom of atomic number $Z$ makes a transition from $m^\text{th}$ to $n^\text{th}$ orbit ($m>n$) is, \begin{align} \label{eaa:eqn:1} \frac{1}{\lambda}=RZ^2\left[\frac{1}{n^2}-\frac{1}{m^2}\right]. \end{align} In the Balmer series, first spectral line corresponds to transition from $m=3$ to $n=1$ and second spectral line corresponds to transition from $m=4$ to $n=2$. Thus, for hydrogen ($Z=1$), \begin{align} \label{eaa:eqn:2} \frac{1}{6561}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5R}{36}, \end{align} and for singly ionized helium ($Z=2$), \begin{align} \label{eaa:eqn:3} \frac{1}{\lambda}=R\times 2^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3R}{4}. \end{align} Divide above equations to get $\lambda={1215}\;{Angstrom}$.