Answer to Question #118771 in Atomic and Nuclear Physics for Jess

Question

The bound electron in a hydrogen atom starts off in an excited state and releases a photon with a wavelength of 95.0 nm as it transitions to the ground state.

What orbital (n) did the electron start in?

Answer 1

Wavelength of photon is $\lambda=95nm$ .

Let us consider the electron was in $n=n_1$ orbital and transits to $n=1$.

Now, from Bohr's principle of hydrogen atom,we know that the amount of energy released by electron in transition is equal to energy of the emitted photon, thus we get

$$\Delta=\frac{hc}{\lambda}=13.6\bigg(\frac{1}{1^2}-\frac{1}{n_1^2}\bigg)eV\\ \implies \frac{1240eV\cdot nm}{\lambda}=13.6\bigg(\frac{1}{1^2}-\frac{1}{n_1^2}\bigg)eV\\ \implies \frac{1240eV\cdot nm}{95nm}=13.6\bigg(\frac{1}{1^2}-\frac{1}{n_1^2}\bigg)eV\\ \implies \frac{1}{n_1^2}=1-0.9597=0.0403‬\\\implies n_1=\sqrt{\frac{1}{0.0403‬}}=4.98$$

But,$n$ is always integer hence,$n_1=5$