Question
Explain the effect of electric potential (potential difference) on photoelectric current and derive the definition and formula of stopping potential.
✓
Answer
→Initially, plate C is illuminated with light keeping plate A at positive potential with respect to plate C and keeping the frequency (v) and intensity (I) of incident light constant. Due to the positive potential of $A$, some of the emitted photo electrons are attracted towards, plate A and form a photocurrent.
→Now if the magnitude (value) of the positive potential of A is increased then more and more emitted electrons are attracted to A and form more photocurrent. Thus, photocurrent increases with voltage.
→After some time, all the electrons emitted from plate C are collected by plate A and form the maximum photocurrent in the circuit. This value of current is saturated (maximum), meaning that on increasing the potential no longer increase the photocurrent. This maximum value of the photocurrent is called saturation current.
• Stopping Potential :
→Now apply a negative potential to the plate A with respect to the plate $C$.
→By doing this the electrons emitted from C will be repelled by A , so the electrons that have enough energy to overcome the repulsion of A will reach A and form a current in the circuit, so the photocurrent will decrease.
→Now if the negative potential of A is increased i.e. made more negative, more and more electrons will be repelled and the number of electrons falling on A will decrease, so the photocurrent will also decrease rapidly.
→For some minimum value of the negative potential of A , all electrons arriving at A will be stopped. i.e. the photocurrent in the circuit will be zero. This minimum value of negative potential of A is called the stopping potential or cut-off voltage.
→Each electron emitted from C has a different energy (kinetic energy), so if the electron with the maximum kinetic energy is stopped by the repulsion of the plate A , all the electrons are stopped and the photocurrent in the circuit is stopped.
→Suppose, maximum kinetic energy of electron is $K _{\max }$.
→The energy provided by the repulsion of A to stop this electron $=e V_0$
→Thus, photocurrent stops if $K _{\max }=e V_0$.
→Here the voltage $V _0$ will be the stopping potential or cut-off voltage.
→Now keeping the frequency of the incident light constant and increasing its intensity (i.e. intensity $I_2$ then $I _3$, where $I _3> I _2> I _1$ ) a graph of photocurrent versus collector plate can be drawn.
→It is clear from this graph that by increasing the intensity of light the value of the maximum saturated photocurrent increases, So $I =\frac{q}{t}=$ $\frac{n e}{t}$ as the number of electrons emitted in 1 sec increases by $\left(\frac{n}{t}\right)$, but the value of the stopping potential does not change, so it can be said from
$K _{\max }=e V_0$, the maximum kinetic energy of emitted photoelectrons does not depend on the intensity of incident light.
$K _{\max }$ does not depend on the intensity of incident radiation.
→Now if the magnitude (value) of the positive potential of A is increased then more and more emitted electrons are attracted to A and form more photocurrent. Thus, photocurrent increases with voltage.
→After some time, all the electrons emitted from plate C are collected by plate A and form the maximum photocurrent in the circuit. This value of current is saturated (maximum), meaning that on increasing the potential no longer increase the photocurrent. This maximum value of the photocurrent is called saturation current.
• Stopping Potential :
→Now apply a negative potential to the plate A with respect to the plate $C$.
→By doing this the electrons emitted from C will be repelled by A , so the electrons that have enough energy to overcome the repulsion of A will reach A and form a current in the circuit, so the photocurrent will decrease.
→Now if the negative potential of A is increased i.e. made more negative, more and more electrons will be repelled and the number of electrons falling on A will decrease, so the photocurrent will also decrease rapidly.
→For some minimum value of the negative potential of A , all electrons arriving at A will be stopped. i.e. the photocurrent in the circuit will be zero. This minimum value of negative potential of A is called the stopping potential or cut-off voltage.
→Each electron emitted from C has a different energy (kinetic energy), so if the electron with the maximum kinetic energy is stopped by the repulsion of the plate A , all the electrons are stopped and the photocurrent in the circuit is stopped.
→Suppose, maximum kinetic energy of electron is $K _{\max }$.
→The energy provided by the repulsion of A to stop this electron $=e V_0$
→Thus, photocurrent stops if $K _{\max }=e V_0$.
→Here the voltage $V _0$ will be the stopping potential or cut-off voltage.
→Now keeping the frequency of the incident light constant and increasing its intensity (i.e. intensity $I_2$ then $I _3$, where $I _3> I _2> I _1$ ) a graph of photocurrent versus collector plate can be drawn.
→It is clear from this graph that by increasing the intensity of light the value of the maximum saturated photocurrent increases, So $I =\frac{q}{t}=$ $\frac{n e}{t}$ as the number of electrons emitted in 1 sec increases by $\left(\frac{n}{t}\right)$, but the value of the stopping potential does not change, so it can be said from
$K _{\max }=e V_0$, the maximum kinetic energy of emitted photoelectrons does not depend on the intensity of incident light.
$K _{\max }$ does not depend on the intensity of incident radiation.
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