Physics - Modern Physics Question with Solution | TestHub

Wavelength of photons released during de-excitation is given by,

$\frac{1}{\lambda }=R\left[\frac{1}{{\displaystyle {\left(n_1\right)}^2}}-\frac{1}{{\displaystyle {\left(n_2\right)}^2}}\right]$.

Clearly, shortest wavelength will correspond to highest energy difference in de-excitation, which in the given figure will correspond to transition from $n=3$ to $n=1.$