Question

Find the ratio of longest wavelength to the shortest wavelength of the Balmer series of hydrogen spectrum.

• A. $1.8$
• B. $2.4$
• C. $1.5$
• D. $1.2$

Answer 1

The correct option is A $1.8$

Wavelength of emitted lines in Hydrogen spectrum is given by
$\frac{1}{\lambda }=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

The longest wavelength of Balmer series corresponds to the transition from $n_i=3to{n}_f=2$


$\frac{1}{{\lambda }_1}=R(\frac{1}{4}-\frac{1}{9})=\frac{5R}{36}$

${\lambda }_1=\frac{36}{5R}$

The shortest wavelength of Balmer series corresponds to the transition from $n_i=\infty to{n}_f=2$

$\frac{1}{{\lambda }_s}=R(\frac{1}{4}-\frac{1}{\infty })=\frac{R}{4}$

${\lambda }_s=\frac{4}{R}$

$\frac{{\lambda }_1}{{\lambda }_s}=\frac{36}{20}=1.8$

Hence, option$\left(a\right)$ is correct.