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 refered to as Lyman and Balmer series.
For Lyman Series wavelength will be longest when energy the electron has transition from $n=2$ to $n=1$ level.
For Balmer Series wavelength will be longest when energy the electron has transition from $n=3$ to $n=2$ level.
$E_2-E_1\propto (-\frac{1}{1^2}+\frac{1}{2^2})=\frac{3}{4}$
$E_3-E_2\propto (-\frac{1}{3^2}+\frac{1}{2^2})=\frac{5}{36}$
$\frac{E_2-E_1}{E_3-E_2}=\frac{\frac{3}{4}}{\frac{5}{36}}=\frac{27}{5}$
Hence, $\frac{{\lambda }_L}{{\lambda }_B}=\frac{5}{27}$ as $E\propto \frac{1}{\lambda }$