Question

The shortest wavelength in Lyman series is $91.2nm$. The longest wavelength of the series is

• A. $121.6nm$
• B. $182.4nm$
• C. $243.4nm$
• D. $364.8nm$

Answer 1

Answer: (A)

The wavelength $(\lambda )$ of lines is given by
$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$
For Lyman series, the shortest wavelength is for
$n=\infty$ and longest is for $n=2$
$\therefore \frac{1}{{\lambda }_S}=R\left(\frac{1}{1^2}\right)\dots$ (i)
$\frac{1}{{\lambda }_L}=R\left(\frac{1}{1}-\frac{1}{2^2}\right)=\frac{3}{4}R\dots$ (ii)
Dividing Eq. (ii) by Eq. (i), we get
$\frac{{\lambda }_L}{{\lambda }_S}=\frac{4}{3}$
Given, ${\lambda }_S=91.2nm$
$\Rightarrow {\lambda }_L=91.2\times \frac{4}{3}$
$=121.6nm$