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

Answer 1

Answer: (B)

$\frac{5}{27}$

Answer 2

The wavelength of a spectral line in the
Lyman series is
$\frac{1}{\lambda_L}=R(\frac{1}{1^2}-\frac{1}{n^2}\bigg),n=2,3,4, .......$
and that in the Balmer series is
$\frac{1}{\lambda_{B}}=R(\frac{1}{2^2}-\frac{1}{n^2}), n=3,4,5,.....$
For the longest wavelength in the Lyman series,
n = 2
$\therefore\quad\frac{1}{\lambda_{L}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=R\left(\frac{1}{1}-\frac{1}{4}\right)=R\left(\frac{4-1}{4}\right)=\frac{3R}{4}$
or $\lambda_L =\frac{4}{3R}$
For the longest wavelength in the Balmer series,
n=3
$\therefore\quad\frac{1}{\lambda_{B}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{9-4}{36}\right)=\frac{5R}{36}$
or $\lambda_B= \frac{36}{5R}$
Thus , $\frac{\lambda_L}{\lambda_B}=\frac{\frac{4}{3R}}{\frac{36}{5R}}=\frac{4}{3R} \times \frac{5R}{36}=\frac{5}{27}$