Physics - Modern Physics Question with Solution | TestHub

Emission of a photon occurs when an electron transitions from a higher energy level to a lower energy level. Thus, transitions from $n=1$ to $n=2$ and $n=2$ to $n=6$ represent absorption, not emission.

For the remaining transitions, the energy of the emitted photon is given by $\Delta E=13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\text{ eV}$.

For $n=2$ to $n=1$:
$\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6\times \frac{3}{4}=10.2\text{ eV}$

For $n=6$ to $n=2$:
$\Delta E=13.6\left(\frac{1}{2^2}-\frac{1}{6^2}\right)=13.6\times \frac{8}{36}\approx 3.02\text{ eV}$

Since frequency $\nu =\frac{\Delta E}{h}$, the transition with the highest energy difference emits the photon of highest frequency. Therefore, the transition from $n=2$ to $n=1$ emits photons of highest frequency.

Answer: $\mathrm{n}=2$ to $\mathrm{n}=1$