The frequency of the light emitted when an electron comes down from $n=4$ to $n=2$ level in hydrogen atom is…
The frequency of the light emitted when an electron comes down from $n=4$ to $n=2$ level in hydrogen atom is $\frac{3}{7}$ times of the following transition of the Li atom
4 to 3
4 to 1
3 to 2
None of these
Solution
We have
$\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
$\begin{aligned} & \Rightarrow \frac{\mathrm{hc}}{\lambda}=\mathrm{hc}_{\mathrm{H}} \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\ & \Rightarrow \quad \mathrm{E}_{\mathrm{n}_2 \rightarrow \mathrm{n}_1}=13.6 \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \mathrm{eV}\end{aligned}$
For hydrogen
$E_{4 \rightarrow 2}=13.6 \times 1^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$
$=13.6\left(\frac{1}{4}-\frac{1}{16}\right)=13.6\left(\frac{4-1}{16}\right)=\frac{13.6 \times 3}{16}=2.55 \mathrm{eV}$
For Lithium
$\mathrm{E}_{\mathrm{n}_2 \rightarrow \mathrm{n}_1}=\frac{3}{7} \times 2.55 \mathrm{eV}=1.09 \mathrm{eV}$
So, $13.6 \times 3^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)=1.09$
None of the value of $n_2$ and $n_1$ satisfy the above equation. So this a none.