Question

The ratio of the wavelength of the last line of the Balmer series to last line of Lyman series is

• A. $1:2$
• B. $2:1$
• C. $4:1$
• D. $1:4$

Answer 1

Answer: (C)

Wavelength of spectral line in hydrogen atom $(z=1)$ given as
$\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\dots \text{ (i) }[\because Z=1]$
For last line of Lyman series,
$n_1=1\text{ and }{n}_2=\infty$
$\therefore$ From Eq. (i), we have
$\frac{1}{{\lambda }_L}=R\left(\frac{1}{1^2}-\frac{1}{\infty }\right)=R$
$\Rightarrow {\lambda }_L=\frac{1}{R}\cdots \cdots (ii)$
Similarly, for last line of Balmer series, $n_1=2$ and $n_2=\infty$
$\therefore$ From Eq. (i), we get
$\frac{1}{{\lambda }_B}=R\left(\frac{1}{2^2}-\frac{1}{\infty }\right)=\frac{R}{4}$
$\Rightarrow {\lambda }_B=\frac{4}{R}$
Hence, from Eq. (i) and Eq. (ii), we get
$\frac{{\lambda }_L}{{\lambda }_B}=\frac{1/R}{4/R}=\frac{1}{4}$
$\text{ or }{\lambda }_L:{\lambda }_B=1:4$
$\Rightarrow {\lambda }_B:{\lambda }_L=4:1$