1. Tardigrade
  2. Question
  3. Physics
  4. The shortest wavelength in Lyman series is 912 mathringA. Then the longest wavelength inthe series must be

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

TS EAMCET 2020

Solution:

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}{912 A}=R\left(\frac{1}{1}-\frac{1}{\infty}\right) $
$\Rightarrow \quad \frac{1}{912 \,\mathring{A}}=R(1-0) $
$\Rightarrow R=\frac{1}{912}(\mathring{A})^{-1} \ldots (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}} $
$\Rightarrow \lambda=1216 \mathring{A}$