Question

The ratio of the longest to the shortest wavelengths in Balmer series of hydrogen spectra is-

• A. $\frac{5}{4}$
• B. $\frac{9}{5}$
• C. $\frac{17}{6}$
• D. $\frac{25}{9}$

Answer 1

The correct option is B $\frac{9}{5}$

As we know, $\Delta E=h\nu =\frac{hc}{\lambda }$

$\Rightarrow \Delta E\propto \frac{1}{\lambda }$

The longest wavelength emitted is in the transition , $n_1=2;n_2=3$

$\frac{1}{{\lambda }_{max}}=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}][\because Z=1]$

$\frac{1}{{\lambda }_{max}}=R[\frac{1}{2^2}-\frac{1}{3^2}]$

${\lambda }_{max}=\frac{36}{5R}.......\left(1\right)$

The shortest wavelength emitted is in the transition , $n_1=2;n_2=\infty$

$\frac{1}{{\lambda }_{min}}=R[\frac{1}{2^2}-\frac{1}{\infty }]$

${\lambda }_{min}=\frac{4}{R}.......\left(2\right)$

On dividing (1) and (2) we get,

$\frac{{\lambda }_{max}}{{\lambda }_{min}}=\frac{\left(\frac{36}{5R}\right)}{\left(\frac{4}{R}\right)}=\frac{9}{5}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.