Question

The shortest wavelength in Lyman series is $912\mathring{A}$. Then the longest wavelength inthe series must be

• A. $9120\mathring{A}$
• B. $1824\mathring{A}$
• C. $1216\mathring{A}$
• D. $2432\mathring{A}$

Answer 1

Answer: (C)

For H-atom (Lyman series formula)
$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)$
For shortest wavelength $\left(n_2=\infty \right)$, so
$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right)$
$\Rightarrow \frac{1}{912A}=R\left(\frac{1}{1}-\frac{1}{\infty }\right)$
$\Rightarrow \frac{1}{912\mathring{A}}=R(1-0)$
$\Rightarrow R=\frac{1}{912}(\mathring{A}{)}^{-1}\dots (i)$
For longest wavelength $\left(n_2=2\right)$, so
$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
$\Rightarrow \frac{1}{\lambda }=\frac{1}{912\mathring{A}}\left(\frac{1}{1}-\frac{1}{4}\right)$
$\Rightarrow \frac{1}{\lambda }=\frac{1}{912\mathring{A}}\left(\frac{3}{4}\right)$
$\Rightarrow \frac{1}{\lambda }=\frac{1}{1216\mathring{A}}$
$\Rightarrow \lambda =1216\mathring{A}$