This problem can be solved using the principle of conservation of energy, as applied in the photoelectric effect. The energy of the incident photon is used to overcome the binding energy (ionization energy) of the electron in the hydrogen atom and the remaining energy is converted into the kinetic energy of the ejected electron.

Step-by-step solution:

1. Formulate the energy conservation equation. The energy of the incident photon ($E_{photon}$) is equal to the sum of the ionization energy of the electron from the $n^{th}$ orbital ($IE_n$) and the kinetic energy of the ejected electron ($KE$). $E_{photon} = IE_n + KE$

2. Calculate the energy of the incident photon. The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the radiation. Given:

   • Wavelength, $\lambda = 90$ nm
   • $hc = 1242$ eV nm

   Substituting the values: $E_{photon} = \frac{1242 \text{ eV nm}}{90 \text{ nm}} = 13.8 \text{ eV}$

3. Determine the ionization energy ($IE_n$). The energy of an electron in the $n^{th}$ orbital of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2}$ eV. The ionization energy is the energy required to remove the electron from the $n^{th}$ orbital to infinity ($E_\infty = 0$). $IE_n = E_\infty - E_n = 0 - \left(-\frac{13.6}{n^2}\right) = \frac{13.6}{n^2} \text{ eV}$

4. Solve for the principal quantum number, n. Now substitute the calculated and given values back into the energy conservation equation:

   • $E_{photon} = 13.8$ eV
   • $KE = 10.4$ eV
   • $IE_n = \frac{13.6}{n^2}$ eV

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

   Rearranging the equation to solve for $IE_n$: $IE_n = 13.8 \text{ eV} - 10.4 \text{ eV} = 3.4 \text{ eV}$

   Now, we can find $n$: $\frac{13.6}{n^2} = 3.4$ $n^2 = \frac{13.6}{3.4} = 4$ $n = \sqrt{4} = 2$

   Since the principal quantum number $n$ must be a positive integer, the value of $n$ is 2.

Final Answer:

The value of n is 2.