If an electron in hydrogen atom jumps from an orbit of level $n=3$ to an orbit at level $n=2$, emitted…
If an electron in hydrogen atom jumps from an orbit of level $n=3$ to an orbit at level $n=2$, emitted radiation has a frequency of ( $R=$ Rydberg's constant and $c=$ velocity of light)
$\frac{3 R c}{27}$
$\frac{R c}{25}$
$\frac{8 R c}{9}$
$\frac{5 R c}{36}$
Solution
When an electron jumps from orbit $n_{i}$ to $n_{f}$, then the frequency of emitted photon is
$v=c R\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)$
Here, $n_{i}=3, n_{f}=2$
$\Rightarrow \quad v=c R\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]=\frac{5 R c}{36}$