Question

The shortest and the longest wavelength in Balmer series of hydrogen spectrum are:
$\text{Rydberg constant},R_H=109678c{m}^{-1}$

• A. 911.7 $\stackrel{o}{A}$ and 1215.7 $\stackrel{o}{A}$
• B. 3647 $\stackrel{o}{A}$ and 6565 $\stackrel{o}{A}$
• C. 6565$\stackrel{o}{A}$ and 3647 $\stackrel{o}{A}$
• D. 911.7 $\stackrel{o}{A}$ and 6565 $\stackrel{o}{A}$

Answer 1

The correct option is B 3647 $\stackrel{o}{A}$ and 6565 $\stackrel{o}{A}$

For Balmer series, $n_1=2.$
For the shortest wavelength in Balmer series (i.e., series limit), the energy difference in the two states showing the transition should be maximum, i.e., $n_2=\infty$
So, $\frac{1}{\lambda }=R_H[\frac{1}{2^2}-\frac{1}{{\infty }^2}]$ = $\frac{R_H}{4}$
$\Rightarrow \lambda =\frac{4}{109678}=3.647\times {10}^{-5}cm=3647\stackrel{o}{A}$

For the longest wavelength in Balmer series (i.e., first line), the energy difference between the states showing the transition should be minimum, i.e. $n_2=3$
$\frac{1}{\lambda }=R_H[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{5}{36}{R}_H\Rightarrow \lambda =\frac{36}{5}\times \frac{1}{R_H}=\frac{36}{5\times 109678}=6.565\times {10}^{-5}cm=6565\stackrel{o}{A}$