Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths ${\lambda }_1$ :  ${\lambda }_2$emitted in the two cases is  [CBSE AIPMT 2012]

Here, for wavelength  ${\lambda }_1,n_1=4\text{ and }{n}_2=3$

and for wavelength  ${\lambda }_2,n_1=3\text{ and }{n}_2=2$

As, we know that,  $\frac{hc}{\lambda }=-13.6\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]$

So, for  ${\lambda }_1$

$\Rightarrow \frac{hc}{{\lambda }_1}=-13.6\left[\frac{1}{(4{)}^2}-\frac{1}{(3{)}^2}\right]$

$\Rightarrow \frac{hc}{{\lambda }_1}=13.6\left[\frac{7}{144}\right]$             …(i)

Similarly,  for ${\lambda }_2$

$\Rightarrow \frac{hc}{{\lambda }_2}=-13.6\left[\frac{1}{(3{)}^2}-\frac{1}{(2{)}^2}\right]$

$\Rightarrow \frac{hc}{{\lambda }_2}=13.6\left[\frac{5}{36}\right]$             …(ii)

Hence, from Eqs. (i) and (ii), we get 

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{20}{7}$