Question

The ratio of minimum wavelengths of Lyman and Balmer series will be

• A. $1.25$
• B. $0.25$
• C. $5$
• D. $10$

Answer 1

Answer: (B)

The series end of Lyman series corresponds to transition from $n_i=\infty$ to $n_f=1$, corresponding to the wavelength,
$\frac{1}{{\left({\lambda }_{\min }\right)}_L}=R\left[\frac{1}{1}-\frac{1}{\infty }\right]=R$
$\Rightarrow {\left({\lambda }_{\min }\right)}_L=\frac{1}{R}=91.2nm$
For last line of Balmer series,
$\frac{1}{{\left({\lambda }_{\min }\right)}_B}=R\left[\frac{1}{(2{)}^2}-\frac{1}{(\infty {)}^2}\right]=\frac{R}{4}$
$\Rightarrow {\left({\lambda }_{\min }\right)}_B=\frac{4}{R}=364.8nm$
Dividing equation (i) by (ii), we get,
$\therefore \frac{{\left({\lambda }_{\min }\right)}_L}{{\left({\lambda }_{\min }\right)}_B}=0.25$