Question

The ratio of the maximum to the minimum wavelengths in Balmer series of H spectrum is

• A. 9 : 5
• B. 36 : 5
• C. 27 : 15
• D. 144 : 7

Answer 1

Answer: (A)

Wavelength of Balmer series is given by
$\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)n=3,4,5,......\infty$
For minimum wavelength , $n=\infty \frac{1}{{\lambda }_{min}}=R\left(\frac{1}{4}-\frac{1}{\infty }\right)=\frac{R}{4}$
or ${\lambda }_{min}=\frac{4}{R}$.......(i)
For maximum wavelength, n = 3
$\frac{1}{{\lambda }_{max}}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5R}{36}$
${\lambda }_{max}=\frac{36}{5R}$ .......(ii)
$\therefore \frac{{\lambda }_{max}}{{\lambda }_{min}}=\frac{36}{5R}\times \frac{R}{4}=\frac{9}{5}$