Question

What is the frequency of photon emitted during a transition from n=4 to n=2 in H atom?

Answer 1

6.17 x 10^14 Hz

Answer 2

The energy of a photon emitted during an electronic transition is given by $\Delta E = h\nu$. For a hydrogen atom, the energy difference between two levels $n_2$ and $n_1$ is $\Delta E = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. The frequency is then $\nu = \frac{R_H}{h} \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R_\infty \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, where $R_\infty \approx 3.29 \times 10^{15} \text{ Hz}$. For the transition from $n=4$ to $n=2$: $\nu = (3.29 \times 10^{15} \text{ Hz}) \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = (3.29 \times 10^{15} \text{ Hz}) \left( \frac{3}{16} \right) \approx 6.17 \times 10^{14} \text{ Hz}$