Experimental Study of Photoelectric Effect: Methods, Observations & Explanation

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The process of emission of photoelectrons from a metal surface when a light beam of a suitable frequency is incident on it, is called photoelectric effect. The emitted electrons are the photoelectrons and the current that is produced is known as the photoelectric current. There are several terms connected to photoelectric effects which are, free electrons, electron emission, photoelectric emission, work function, cut-off potential, cut-off frequency and cut-off wavelength. 

Key Terms: Photoelectric Effect, Einstein Photoelectric Equation, Photoelectric Emission, Photons, Electrons, Ammeter


Description of the Experiment

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The aim of the experiment is to review the emission of electrons by light. We also attempt to measure the energy of the electrons emitted within the process. Additionally to the present, we'll also observe the relation of those electrons with the frequency of sunshine used. To review the effect, we use an evacuated beam tube connected during a circuit as shown below.

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Apparatus Used in Experimental study of the photoelectric effect

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Quartz Window

Near one among the plates inside the evacuated tube, there is a small quartz window. The Quartz window has two significant functions – it lets light in and it only lets the Ultra Violet light in. Therefore, by employing a Quartz window, we confirm that light of a selected frequency falls on the metal plate inside the evacuated chamber.

Experimental study of Photoelectric Effect

Experimental Study of Photoelectric Effect

The circuit

We connect a voltmeter across the 2 plates. This is responsible for measuring the electric potential between the plates. Moreover, we have a sensitive galvanometer within the circuit which measures the photocurrent.

The emitter plate-C emits electrons which are then assembled at the collector plate-A. These plates are attached by the electric field to the battery via the commutator.


Initial stage of the Experiment

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In the beginning, let us assume there is zero potential. We open the quartz window and then study the reading of the Voltmeter and therefore the Ammeter. Both will provide a non-zero reading (such as for alkali metals) by proving the occurrence of the Photoelectric effect. As we increase the Voltage and alter it again, we'll get the subsequent observations:


Effect of Intensity

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The number of electrons emitted per second is observed to be directly proportional to the intensity of sunshine. It is due to the electrons emitted from C are attracted to the collector A, the collector A is maintained at a positive potential with reference to emitter C. With a fixed frequency of incident radiation and the accelerating potential, the intensity of light is varied and concluding photoelectric current is measured every time. Hence, the photoelectric current increases linearly with the intensity of incident light.

As we see, this only takes place above a selected value of frequency, referred to as the threshold frequency. Below this threshold frequency, the intensity of light is effectless on the photocurrent. In fact, howsoever high the intensity of light is, there's no photocurrent at all.

Photoelectric Current and Intensity of Light

Photoelectric Current and Intensity of Light

The above graph is between the photoelectric current and the intensity of light in a line when the frequency of light used is beyond a selected minimum threshold value.


Effect of Potential

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Think, if you connect C to a positive terminal and A to a negative terminal. What does one expect will happen to the photocurrent?

Effect of Potential of Photoelectric Current

Effect of Potential of Photoelectric Current

Here, the frequency and intensity of radiation is fixed. Since electrons are charged, if we increase the negative potential at C, more and more electrons will want to flee this region and run to the attractive plate A. Therefore the current should increase. Similarly, if we decrease the negative potential at C, removing electrons will become difficult and therefore the photocurrent will decrease. Hence the utmost current flowing at a given intensity of incoming light is that the saturation current.

As you'll see within the graph, the worth of saturation current is bigger for higher intensities, provided the frequency is above the edge frequency. Imagine you're an electron and you only escaped the metal surface. Now you're merrily accelerating towards A. What if we became mischievous and increased the negative potential at A? You’ll feel a repulsion and consequently you'll lose speed.

What if the potential is extremely strong? You’ll not be ready to escape the metal surface at all! As a result, we call this value of the potential that the photocurrent becomes zero because of the stopping potential or the retarding potential. The increased negative potential of the collector plate, the more effort that an electron has got to make if it wants to flee successfully from the metal surface.

Thus we will get the subsequent relationship between the stopping potential and therefore the photocurrent.


Effect of Frequency

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We see that for higher frequency values like ν3, stopping potential is more negative or greater than the stopping potential for smaller frequencies like ν1. What does this mean? This suggests that there should be a relationship between the frequency and energy.

Effect of Frequency of Incident Radiation

Effect of Frequency of Incident Radiation


Solved Question

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  1. Zero
  2. Less
  3. More
  4. Remains Unchanged

Solution: (3) It will become more.

We know that the energy of photoelectrons increases as we increase the frequency. This means that their kinetic energy will be more. Hence higher frequency means a greater speed of a photoelectron. We also know that λ = c/ν. Hence if the wavelength is increased, the frequency will be decreased and vice-versa. So lesser wavelength means greater frequency and greater speed of the photoelectrons.


Things to Remember

  • The process of emission of photoelectrons from a metal surface when a light beam of a suitable frequency is incident on it, is called photoelectric effect.
  • There are several terms connected to photoelectric effects which are, free electrons, electron emission, photoelectric emission, work function, cut-off potential, cut-off frequency and cut-off wavelength. 
  • Near one among the plates inside the evacuated tube, there is a small quartz window.
  • The number of electrons emitted per second is observed to be directly proportional to the intensity of sunshine.
  • We have a sensitive galvanometer within the circuit which measures the photocurrent.
  • For a given metal (photosensitive material), the photoelectric current is in direct proportion to the intensity of light that is used, above a minimum value of frequency, the threshold frequency. 

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Sample Questions

Ques. What are the apparatus used for the experimental study of photoelectric effect? (2 marks)

Ans. The apparatus used for the experiment are as follows,

Quartz Window- Near one among the plates inside the evacuated tube, there is a small quartz window. The Quartz window has two significant functions – it lets light in and it only lets the Ultra Violet light in. Therefore, by employing a Quartz window, we confirm that light of a selected frequency falls on the metal plate inside the evacuated chamber.

The circuit- We connect a voltmeter across the 2 plates. This is responsible for measuring the electric potential between the plates. Moreover, we have a sensitive galvanometer within the circuit which measures the photocurrent. The emitter plate-C emits electrons which are then assembled at the collector plate-A. These plates are attached by the electric field to the battery via the commutator.

Ques. Estimate the number of photons emitted by the bulb per second. (2 marks)

Ans. P = n1 c/λ

n1 = Pλ / hc = 20 x 5000 x 10-10 / 6.62 x 10-34 x 3 x 108

n1 = 5 x 1019 

This is the number of photons per second.

Ques. Light of a specific frequency v is incident on a metal surface. If the intensity of incident radiation increases, then photoelectric current
a) Decreases b) Increases c) Remains unchanged d) sometimes increases and sometimes decreases (2 marks)

Ans. b) increases

The intensity of light is inversely proportional to the number of photons that are striking the metal per unit time. Again, the number of photons striking the metal surface is inversely proportional to the electrons knocked out of metal per unit time.

Current = charge/ time

Therefore, more the electrons knocked out per unit time, more is the photoelectric current.

Ques. How can the photoelectric effect be observed instantaneously? (2 marks)

Ans. The time of electron emission is 11.4 sec. So the photoelectric emission is not immediate in the problem. It takes around 11.4 sec. In the photoelectric emission, there is a collision between free electron, incident photon and nucleus which lasts for a very short period of time ( 10-9 sec). Therefore, we can say that the photoelectric emission is instantaneous.

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CBSE CLASS XII Related Questions

  • 1.
    The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

      • \( \frac{\alpha}{r} \)
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    • 2.
      A ray of light is incident on a refracting face AB of a prism ABC at an angle of \( 45^\circ \). The ray emerges from face AC and the angle of deviation is \( 15^\circ \). The angle of prism is \( 30^\circ \). Show that the emergent ray is normal to the face AC from which it emerges out. Find the refraction index of the material of the prism.


        • 3.
          A charge \( -6 \, \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \, \mu C \) is moved from point C to point A along the circumference. Calculate the work done on the charge.
          A charge −6µC is placed


            • 4.

              Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \). 


                • 5.
                  A 1 cm segment of a wire lying along the x-axis carries a current of 0.5 A along the \( +x \)-direction. A magnetic field \( \vec{B} = (0.4 \, \text{mT} \hat{j}) + (0.6 \, \text{mT} \hat{k}) \) is switched on. The force acting on the segment is:

                    • \( (2 \hat{i} + 3 \hat{k}) \, \text{mN} \)
                    • \( (-3 \hat{j} + 2 \hat{k}) \, \mu\text{N} \)
                    • \( (6 \hat{j} + 4 \hat{k}) \, \text{mN} \)
                    • \( (-4 \hat{j} + 6 \hat{k}) \, \mu\text{N} \)

                  • 6.
                    Two point charges \( 5 \, \mu C \) and \( -1 \, \mu C \) are placed at points \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \), respectively. An external electric field \( \vec{E} = \frac{A}{r^2} \hat{r} \) where \( A = 3 \times 10^5 \, \text{V m} \) is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.

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