Principal Quantum Number of Hydrogen Atom Ionized by Radiation

To find the principal quantum number $n$ of the hydrogen atom, we apply the conservation of energy during photoionization.

Step 1: Energy of the incident photon The energy $E$ of the electromagnetic radiation of wavelength $\lambda = 90\text{ nm}$ is given by: $E = \frac{hc}{\lambda}$

Given $hc = 1242\text{ eV nm}$, we have: $E = \frac{1242\text{ eV nm}}{90\text{ nm}} = 13.8\text{ eV}$

Step 2: Energy conservation during ionization The energy of the electron in the $n^{\text{th}}$ orbital of a hydrogen atom is given by: $E_n = -\frac{13.6}{n^2}\text{ eV}$

When the incident photon ionizes the atom, the energy conservation equation is: $E_n + E = K_{\text{ejected}}$

where $K_{\text{ejected}} = 10.4\text{ eV}$ is the kinetic energy of the ejected electron.

Step 3: Calculating $n$ Substituting the values into the energy conservation equation: $-\frac{13.6}{n^2} + 13.8 = 10.4$

$-\frac{13.6}{n^2} = 10.4 - 13.8 = -3.4\text{ eV}$

$\frac{13.6}{n^2} = 3.4$

$n^2 = \frac{13.6}{3.4} = 4$

$n = 2$

Thus, the value of $n$ is $2$.