Question

A photon emitted during the de-excitation of electron from a state $n$ to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function $2\ \text{eV}$ in a photocell with a stopping potential of $0.55\ \text{V}$. Obtain the value of the quantum number of the state $n$?

Answer 1

Hint: A photon is emitted when an electron jumps from a higher energy level to a lower energy during the de-excitation. Now, according to Einstein's photoelectric equation, a photon when irradiated on a metallic cathode can liberate electrons from its surface of the energy of the photon is greater than the stopping potential of the electrode.
Formula Used:
Energy of a photon in Hydrogen atom is given by,
${{E}_{n}}=-\dfrac{13.6}{{{n}^{2}}}\ \text{eV}$
Here, $n$ is the level in which electrons reside.
The Einstein’s photoelectric equation can be written as,
${{E}_{photon}}=W+q{{V}_{0}}$
Here, ${{E}_{photon}}$ is energy of incoming photon, $W$ is the work function of photo-cell, $q$ is the electronic charge and ${{V}_{0}}$ is the stopping potential.

Complete answer:
Energy of a photon in Hydrogen atom is given by,
${{E}_{n}}=-\dfrac{13.6}{{{n}^{2}}}\ \text{eV}$
Now, let the electron is initially in some $n$ state and jumps to the first excited state $n=2$. Thus, energy of photon liberated is calculated as,
\[\begin{align}
  & {{E}_{photon}}={{E}_{i}}-{{E}_{f}} \\
 & =\left[ -\dfrac{13.6}{{{n}^{2}}}\ \text{eV} \right]-\left[ -\dfrac{13.6}{4}\ \text{eV} \right] \\
 & =\left( 3.4-\dfrac{13.6}{{{n}^{2}}} \right)\ \text{eV}
\end{align}\]
Now, use Einstein’s photoelectric equation to model the situation when the liberated photon strikes the metallic cathode. It can be modelled as,
${{E}_{photon}}=W+q{{V}_{0}}$
Now, substitute the given values as $W=2\ \text{eV}$ and ${{V}_{0}}=0.55\ \text{eV}$. Now calculate the initial quantum number. The calculations can be shown as,
$\begin{align}
  & \left( 3.4-\dfrac{13.6}{{{n}^{2}}} \right)\ \text{eV}=\left( 2\ \text{eV} \right)+\left( 0.55\ \text{eV} \right) \\
 & 3.4-2.55=\dfrac{13.6}{{{n}^{2}}} \\
 & {{n}^{2}}=\dfrac{13.6}{0.85} \\
 & {{n}^{2}}=16
\end{align}$
This gives $n=4$. Thus, the electron jumps from third excited state to first excited state.

Note:
The energy of an electron in a quantum state of hydrogen is given by ${{E}_{n}}=-\dfrac{13.6}{{{n}^{2}}}\ \text{eV}$. Take care of the negative sign while calculating the energy difference between two energy levels. Also, take care of the units while substituting the values in Einstein's photoelectric equation so that all the quantities on both sides of the inequality are in electron volts.