Question

Ultraviolet radiations of different frequencies ${v_1}$ and ${v_2}$, are incident on two photosensitive materials having work functions ${W_1}$ and ${W_2}$(${W_1}$>${W_2}$) respectively. The kinetic energy of the emitted photoelectrons is the same in both cases. Which one of the two relations will be of the higher frequency?

Answer 1

Hint: In case of photoelectric effect, the cathode is hit by a beam of photons which causes electrons in the cathode to come out of the cathode and go to anode. Potential difference is maintained between cathode and anode. The intensity of the incident light determines the current and the energy of the incident light determines the maximum energy of the electron
Formula used:
$hv = W + E$

Complete answer:
When radiation is hit on the cathode some of the energy of that radiation will be used by the electron to come out of that electrode which is called a work function and the remaining energy is used as a kinetic energy to reach the anode. When an electron reaches anode automatically current is generated. The amount of current produced depends on the intensity of the radiation.
We can assume that each photon will bring one electron out of the metal. Hence ‘n’ photons will bring ‘n’ electrons out of the metal.
Now some energy from the photon is used as a work function to remove electrons from the metal and the rest will be the remaining energy. Hence the maximum kinetic energy of an electron will be
$E = hv - W$
Where $W$ is the work function and ‘h’ is the planck's constant and $v$ is the frequency of the light and ‘E’ is the maximum kinetic energy.
Here two ultraviolet radiations of frequencies ${v_1}$ and ${v_2}$ are hit on two photosensitive materials having work functions ${W_1}$ and ${W_2}$
so from the above equation we get
$hv = W + E$
In the question it is given that the kinetic energy is the same in both cases. So for the first radiation we have
$h{v_1} = {W_1} + E$
for the second radiation we have
$h{v_2} = {W_2} + E$
It is given that ${W_1}$>${W_2}$ so ${v_1}$>${v_2}$
So the frequency of first radiation will be more than the second one.

Note:
This maximum kinetic energy determines the stopping potential. It means the potential maintained at anode relative to cathode such that no electron reaches the anode. It doesn’t matter if we increase the number of photons, but as long as the photon energy is less than the required work function, the electron will not emit from the metal. If we apply the stopping potential at the anode then no electron will reach the anode and there will be no photoelectric current.