Physics - Modern Physics Question with Solution | TestHub

$\_\_\_\_\_\mathit{ }n=4$

$\_\_\_\_\_\mathit{ }n=3$

$\_\_\_\_\_\mathit{ }n=2$

$\_\_\_\_\_\mathit{ }n=1$

The wavelength of the light released when an electron jumps from higher energy state $\left(n_2\right)$ to lower energy state $\left(n_1\right)$ is $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{{\displaystyle 1}}{{\displaystyle n_2^2}}\right)$, here, $\lambda$ is wavelength, $R$ is Rydberg Constant. 

When transition is from second excited state to first excited state, 
$n=3\to n=2$

$\frac{1}{{\lambda }_{\mathit{0}}}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) ...\left(1\right)$  

When transition is from third excited state to second orbit

$n=4\to n=2$

$\frac{1}{\left({\displaystyle \frac{20}{x}}{\lambda }_0\right)}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) ...\left(2\right)$

Dividing $\left(2\right) \mathrm{by} \left(1\right)$

$\Rightarrow \frac{x}{20}=\frac{\frac{1}{2^2}-\frac{1}{4^2}}{\frac{1}{2^2}-\frac{1}{3^2}}=\frac{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{16}}}{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{9}}}=\frac{{\displaystyle \frac{3}{16}}}{{\displaystyle \frac{5}{36}}}=\frac{27}{20}$

$\Rightarrow x=27$