If an electron in hydrogen atom jumps from $3^{\text {rd }}$ orbit to $2^{\text {nd }}$ orbit, it emits a…
If an electron in hydrogen atom jumps from $3^{\text {rd }}$ orbit to $2^{\text {nd }}$ orbit, it emits a photon of wavelength ' $\lambda$ '. When it jumps from $4^{\text {th }}$ orbit to $3^{\text {rd }}$ orbit, the corresponding wavelength of the photon will be
$\frac{20}{13} \lambda$
$\frac{20}{7} \lambda$
$\frac{9}{16} \lambda$
$\frac{16}{25} \lambda$
Solution
Rydberg's relation is given by
$\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
For hydrogen, $\mathrm{Z}=1$.
Given, for $3^{\text {rd }}$ to $2^{\text {nd }}$ orbit transition
And for $4^{\text {th }}$ to $3^{\text {rd }}$ orbit transition
Taking ratio of eqn (1) \& (2)
$\begin{aligned}
& \frac{\lambda_n}{\lambda}=\frac{5}{36} \times \frac{(16 \times 9)}{7}=\frac{20}{7} \\
& \Rightarrow \lambda_n=\frac{20}{7} \lambda
\end{aligned}$
And for $4^{\text {th }}$ to $3^{\text {rd }}$ orbit transition
Taking ratio of eqn (1) \& (2)
$\begin{aligned}
& \frac{\lambda_n}{\lambda}=\frac{5}{36} \times \frac{(16 \times 9)}{7}=\frac{20}{7} \\
& \Rightarrow \lambda_n=\frac{20}{7} \lambda
\end{aligned}$