Question

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

• A. $\frac{5}{27}$
• B. $\frac{4}{9}$
• C. $\frac{9}{4}$
• D. $\frac{27}{5}$

Answer 1

The correct option is A $\frac{5}{27}$

Transition to $n=1$ and $n=2$ are referred to as Lyman and Balmer series respectively.

For Lyman series, wavelength will be longest when the electron has a transition from $n=2$ to $n=1$ level.
For Balmer series wavelength will be longest when electron has a transition from $n=3$ to $n=2$ level.
as $E\propto \frac{1}{\lambda }$
$\frac{1}{{\lambda }_L}\propto (\frac{1}{1^2}-\frac{1}{2^2})=\frac{3}{4}$
$\frac{1}{{\lambda }_B}\propto (-\frac{1}{3^2}+\frac{1}{2^2})=\frac{5}{36}$
$\frac{\frac{1}{{\lambda }_L}}{\frac{1}{{\lambda }_B}}=\frac{\frac{3}{4}}{\frac{5}{36}}=\frac{27}{5}$
Hence, $\frac{{\lambda }_L}{{\lambda }_B}=\frac{5}{27}$