The Photoelectric Effect

The photoelectric effect is a fundamental phenomenon in physics where electrons are ejected from a material’s surface when exposed to light of a certain frequency. Discovered by Heinrich Hertz and later explained by Albert Einstein, this effect demonstrated that light has particle-like properties and supported the quantum theory of light. In everyday life, the principles of the photoelectric effect are integral to various technologies. For instance, solar panels harness sunlight to generate electricity, leveraging the effect to convert light energy into electrical energy. Similarly, photoelectric sensors used in automatic doors, security systems, and even some types of cameras rely on this phenomenon to detect light and trigger responses. In this article, we will discuss the concept of the Photoelectric effect, the application of the photoelectric effect, graphs related to the photoelectric effect and solved examples for better understanding.

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This Story also Contains

1. Photoelectric Effect
2. Applications of Photoelectric Effect
3. Graphs Related to the Photoelectric Effect
4. Solved Examples Based on the Photoelectric Effect
5. Summary

Photoelectric Effect

The phenomenon of the Photoelectric effect was first introduced by Wilhelm Ludwig Franz Hallwachs in 1887 and its experimental verification was confirmed by Heinrich Rudolf Hertz. They observed that when a metallic surface is irradiated by monochromatic light of proper frequency, electrons are emitted from it. This phenomenon of ejection of electrons is called the Photoelectric effect. Now the question arises what is the Photoelectric effect

The photoelectric effect is the process that involves the release or rejection of electrons from the surface of materials (this material is generally a metal) when light falls on them. This concept makes us comfortable understanding the quantum nature of electrons and light.

The electrons ejected during the photoelectric effect were called photoelectrons. There is one condition for the photoelectric effect which is very important for photoemission to take place, energy of incident light photons should be greater than or equal to the work function of the metal. Now what is work function?

Work function $(\phi)$ is defined as the minimum quantity of energy which is required to remove an electron to infinity from the surface of a given solid, usually a metal.

Now on the basis of work function $(\phi)$, we can define two related quantities which are Threshold frequency and Threshold wavelength. Now as we know the energy of a photon is given by

$E=h \nu=\frac{h c}{\lambda}$

Now the frequency corresponding to the energy equal to the work function is called Threshold frequency and similarly, the wavelength corresponding to the work function is Threshold wavelength.

$
\phi=h \nu_{t h} \quad \nu_{t h}=\frac{\phi}{h}
$

Similarly $\phi=\frac{h c}{\lambda_{t h}} \quad \lambda_{t h}=\frac{h c}{\phi}$

Now, let us understand an experiment which was performed by Heinrich Rudolf Hertz. For this let us consider the given setup

In this experiment setup, an evacuated glass tube is there. Two zinc plates C and A are enclosed. Plates A acts as an anode and C acts as a photosensitive plate. Two plates are connected to a battery and ammeter as shown. If the radiation is incident on the plate C through a quartz window, electrons are ejected out of the plate and current flows in the circuit this is known as photocurrent. Plate A can be maintained at the desired potential (+ve or – ve) with respect to plate C.

Applications of Photoelectric Effect

• This phenomenon is used to generate electricity in Solar Panels.
• We come across many sensors in our day-to-day lives. A few sensors are also working in the Photoelectric effect.
• It is also used in digital cameras because they have photoelectric sensors.

Note

• In case of Threshold frequency - If incident frequency $v<\nu_0$ . No photoelectron emission. The minimum frequency of incident radiation to eject electron is threshold frequency $\left(\nu_0\right)$
• In the case of Threshold Wavelength - If $\lambda>\lambda_0$ No photoelectron emission. The maximum wavelength of incident radiation required to eject the electron is the Threshold Wavelength $\left(\lambda_0\right)$
• Work function

  $\begin{aligned} & h=\text { Planck's constant } \\ & \nu_0=\text { threshold frequency }\end{aligned}$

  Energy is used to overcome the surface barrier and come out of the metal surface.

  $\phi=h \nu_0$

• Kinetic Energy of Photo Electrons

  $m \rightarrow$ mass of photoelectron

  The remaining part of the energy is used in gaining a velocity $v$ to the emitted photoelectron

  $k_{\max }=\frac{1}{2} m v_{\max }^2$

• Conservation of energy

• The conservation of energy in the photoelectric effect is a fundamental principle that describes how the energy of incident photons is used to liberate electrons from a material. According to Einstein's photoelectric equation, the energy of a photon E_(photon) is used to overcome the work function $(\Phi)$ of the material and provide the ejected electron with kinetic energy (KE).

  

  $
  \begin{aligned}
  & h \nu=\phi_0+\frac{1}{2} m v_{\max }^2 \\
  & h \nu=h \nu_0+\frac{1}{2} m v_{\max }^2 \\
  & h\left(\nu-\nu_0\right)=\frac{1}{2} m v_{\max }^2
  \end{aligned}
  $
  where, $h-$ Planck's constant, $\nu-$ Frequency, $\nu_0$-threshold frequency, $\phi_0-$ work function

Graphs Related to the Photoelectric Effect

Graphs related to the photoelectric effect typically illustrate the relationship between various parameters such as the intensity of light, frequency of light, and the kinetic energy of emitted electrons. Here are the key graphs associated with the photoelectric effect. Before giving the variation in the graph we should define some important terminologies which are used while plotting the graph.

1. Stopping Potential

The negative potential of the collector plate at which the photoelectric current becomes zero is called the stopping potential or cut-off potential. Stopping potential is the value of retarding potential difference between two plates which is just sufficient to stop the most energetic photoelectrons emitted. It is denoted by $V_o$.

We need to equate the maximum kinetic energy Kmax of the photo-electron (having charge e) to the stopping potential $V_o$.

We know that,
Electric potential energy = Potential Difference×Charge
So, $\begin{aligned} & U=V_0 \times Q \\ & U=K_{\max } \\ & \therefore\left|V_0 \times e\right|=K_{\max } \\ & \Rightarrow K_{\max }=\left|e V_0\right|\end{aligned}$

By using the previous concept, we can write that

$h \nu=h \nu_o+K \cdot E_{\cdot(\max )}$

Now since $\mathrm{K} \cdot \mathrm{E}_{\cdot \max }=\mathrm{eV}_0$, So we can write that

$\begin{gathered}\mathrm{eV}_{\mathrm{s}}=\mathrm{h}\left(\nu-\nu_{\mathrm{o}}\right) \\ \text { or } \\ \mathrm{V}_{\mathrm{s}}=\frac{\mathrm{h}}{\mathrm{e}}\left(\nu-\nu_{\mathrm{o}}\right)\end{gathered}$

The above graph shows the variation between the stopping potential and frequency.

2. Saturation Current

The photoelectric current attains a saturation value and does not increase further for any increase in the positive potential. It means that this photoelectric current is the saturation current even we are increasing the value of the positive potential. Now let us discuss the variations one by one in detail.

1. Variation of photocurrent with intensity

2. Variation of photoelectric current with potential and intensity

3. Effects of frequency of incident light on the stopping potential

4. Variation of Kinetic energy with frequency

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Solved Examples Based on the Photoelectric Effect

Example 1: The work function for $\mathrm{Al}, \mathrm{K}$ and Pt is $4.28 \mathrm{eV}, 2.30 \mathrm{eV}$ and 5.65 eV respectively. Their respective threshold frequencies would be

1) $P t>A l>K$
2) $A l>P t>K$
3) $K>A l>P t$
4) $A l>K>P t$

Solution

Work function, $\phi_0=\mathrm{h} \nu_0$
Where $\nu_0=$ Threshold frequency
So, $\phi_0 \propto \nu_0$
Hence $\mathrm{Pt}>\mathrm{Al}>\mathrm{K}

Hence, the answer is the option (1).

Example 2: A and B are two metals with threshold frequencies $1.8 \times 10^{14} \mathrm{~Hz}$ and $2.2 \times 10^{14} \mathrm{~Hz}$. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted in:

(Take $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$ )

1) B alone

2) A alone

3) neither A nor B

4) both A and B

Solution:

$\begin{aligned} & \phi_{0_{\mathrm{A}}}=\frac{\mathrm{hv}_0}{\mathrm{e}} \mathrm{eV}=\frac{\left(6.6 \times 10^{-34}\right) \times\left(1.8 \times 10^{14}\right)}{1.6 \times 10^{-19}} \mathrm{eV}=0.74 \mathrm{eV} \\ & \phi_{0_{\mathrm{B}}}=\frac{\left(6.6 \times 10^{-34}\right) \times\left(2.2 \times 10^{14}\right)}{1.6 \times 10^{-19}} \mathrm{eV}=0.91 \mathrm{eV}\end{aligned}$

Since the incident energy 0.825 eV is greater than 0.74 eV and less than 0.91 eV, so photoelectrons are emitted from metal A only.

Hence, the answer is the option (2).

Example 3: When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively:
1) N and 2T

2) 2N and T

3) 2N and 2T

4) N and T

Solution:

The number of photoelectrons ejected is directly proportional to the intensity of incident light. Maximum kinetic energy is independent of the intensity of incident light but depends upon the frequency of light.

Hence, the answer is the option (2).

Example 4: Sodium and copper have work functions of 2.3 eV and 4.5 eV respectively. Then, the ratio of their threshold wavelengths is nearest to:

1) 1:2

2) 2:1

3) 1:4

4) 4:1

Solution:

Work function, $\mathrm{W}_0=h \mathrm{v}_0$
$
\begin{aligned}
& \text { For sodium, } \mathrm{W}_{01}=\mathrm{hv}_{01}=\frac{\mathrm{hc}}{\lambda_0} \\
& =\frac{2.3 \mathrm{eV}}{\mathrm{h}}=\mathrm{v}_{01}=\frac{\mathrm{c}}{\lambda_{01}}
\end{aligned}
$

For copper, $\mathrm{W}_{02}=\mathrm{hv}_{02}$
$
\begin{aligned}
& =\frac{4.5 \mathrm{eV}}{\mathrm{h}}=\mathrm{v}_{02}=\frac{\mathrm{c}}{\lambda_{02}} \\
& =\frac{\lambda_{01}}{\lambda_{02}}=\frac{\mathrm{ch}}{2.3} \times \frac{4.5}{\mathrm{ch}}=\frac{2}{1} \\
& \Rightarrow \lambda_{01}: \lambda_{02}=2: 1
\end{aligned}
$

Hence, the answer is the option (2).

Summary

The photoelectric effect, discovered by Heinrich Hertz and explained by Albert Einstein, describes how electrons are emitted from a material's surface when illuminated by light of a specific frequency. This phenomenon, essential for understanding the quantum nature of light, is foundational to technologies like solar panels and photoelectric sensors in everyday devices. The key aspects include the work function, which is the energy required to release electrons, and concepts like threshold frequency and wavelength. Graphs related to the photoelectric effect illustrate how the stopping potential, photocurrent and kinetic energy of emitted electrons vary with light frequency and intensity.