Question

The photon emitted during the de-excitation from the first excited level to the ground state of hydrogen atom is used to irradiate a photo cathode of a photocell, in which stopping potential of $5V$ is used. Calculate the work function of the cathode used.

Answer 1

In photoelectric effect, stopping potential is directly proportional to the maximum kinetic energy of the emitted electron. This maximum kinetic energy is further dependent on the frequency of incident light as well as work function of the cathode. Combining both these facts, we can arrive at the work function of the cathode used.

Formula used:
$1)e{{V}_{s}}=K{{E}_{\max }}$
$2)K{{E}_{\max }}=h\nu -\phi$
$3){{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}eV$

Complete step by step answer:
In photoelectric effect, stopping potential is directly proportional to the maximum kinetic energy of the emitted electron. Mathematically, this can be expressed as:
$e{{V}_{s}}=K{{E}_{\max }}$
where
$e$ is the electronic charge
${{V}_{s}}$ is the stopping potential
$K{{E}_{\max }}$ is the maximum kinetic energy of the emitted electron
Let this be equation 1.
Further, maximum kinetic energy of the emitted electron is dependent on frequency of the incident light as well as the work function of the cathode used as follows:
$K{{E}_{\max }}=h\nu -\phi$
where
$K{{E}_{\max }}$ is the maximum kinetic energy of the emitted electron
$h$ is the Planck’s constant
$\nu$ is the frequency of incident light
$\phi$ is the work function of cathode
Let this be equation 2.
Combining both equation 2 and equation 1, we have
$e{{V}_{s}}=h\nu -\phi$
Let this be equation 3.
Coming to our question, we are provided that the photon emitted during the de-excitation from the first excited level to the ground state of hydrogen atom is used to irradiate a photo cathode of a photocell, in which stopping potential of $5V$ is used. We are required to calculate the work function of the cathode used.
Clearly,
$h\nu ={{E}_{2}}-{{E}_{1}}=\dfrac{-13.6}{{{2}^{2}}}-\dfrac{-13.6}{{{1}^{2}}}=10.2eV$
where
$h\nu$ is the energy required for de-excitation of electron from the first excited state to the ground state
${{E}_{2}}$ is the energy of electron in the first excited state
${{E}_{1}}$ is the energy of electron in the ground state
Let this be equation 4.
Substituting equation 4 as well as the given value of stopping potential in equation 3, we have
$e{{V}_{s}}=h\nu -\phi \Rightarrow e\times 5V=10.2eV-\phi \Rightarrow \phi =5.2eV$
Let this be equation 5.
Therefore, from equation 5, we can conclude that the work function of the cathode used is equal to $5.2eV$.

Note:
Students need to remember that energy of electron in ${{n}^{th}}$ excited state is given by
${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}eV$
where
$n=1,2,3,....etc$ are the different energy states.
Here it is important to note that this energy is expressed in $eV$. This unit can further be noticed in the calculations of equation 5 to express the work function of the cathode used.