U4AOS1 Topic 20: Photoelectric Effect

The photoelectric effect is a phenomenon in which electrons are ejected from a material's surface when it absorbs light. This effect provided crucial evidence for the quantum nature of light and helped establish quantum mechanics. Albert Einstein's explanation of the photoelectric effect earned him the Nobel Prize in Physics in 1921.

Historical Background

The photoelectric effect was first observed by Heinrich Hertz in 1887. Hertz noted that ultraviolet light facilitated the passage of electric sparks between metal electrodes. Wilhelm Hallwachs and Philipp Lenard conducted further experiments, showing that light could indeed cause the emission of electrons from metal surfaces. However, classical wave theories of light could not explain the observations, leading to the need for a new theoretical framework.


The photoelectric effect is where a current is applied in a open circuit. The open end of the circuit has 2 parts:

                                         Cathode: where the electrons gather at the end of the circuit and has light shined on it.

Anode: where the electrons end up after jumping over the open circuit due to the energy provided by the photons.

Key Observations

Several critical observations characterize the photoelectric effect:

  1. Threshold Frequency: For each material, there is a minimum frequency of incident light, called the threshold frequency, below which no electrons are ejected, regardless of light intensity.

  2. Kinetic Energy of Ejected Electrons: The kinetic energy of the ejected electrons depends on the frequency of the incident light and not its intensity. Higher frequency light ejects electrons with greater kinetic energy.

  3. Instantaneous Emission: The emission of electrons occurs almost instantaneously after the light hits the material, suggesting that the energy transfer is not a cumulative process.

Einstein's Explanation

Albert Einstein proposed a quantum theory of light to explain the photoelectric effect. Building on Max Planck's idea of quantized energy, Einstein suggested that light consists of packets of energy called photons. Each photon has an energy E proportional to its frequency fff:

E=hE

where h is Planck's constant (6.626×10−34J s).

Einstein's theory can be summarized in the following points:

  1. Photon Energy: A photon with frequency f has an energy E=hf.

  2. Energy Transfer: When a photon hits the material, it transfers its energy to an electron.

  3. Work Function (ϕ): Each material has a work function, ϕ, which is the minimum energy required to eject an electron from the material.

  4. Kinetic Energy: If the photon's energy exceeds the work function, the excess energy becomes the kinetic energy (K.E.) of the ejected electron:

K.E.=hf−ϕ

Experimental Validation

Robert Millikan conducted experiments to validate Einstein's theory. He measured the kinetic energy of ejected electrons as a function of light frequency and found a linear relationship, confirming the equation K.E.=hf−ϕ

Detailed Analysis

  1. Threshold Frequency: The threshold frequency f0f_0f0​ is the minimum frequency needed to overcome the work function ϕ:

ϕ=hf

Below this frequency, photons do not have enough energy to eject electrons.

  1. Intensity and Electron Emission: Increasing the intensity of light increases the number of photons but not their individual energy. Hence, more electrons are ejected, but their kinetic energy remains the same if the frequency is constant.

  2. Stopping Potential: In experiments, a stopping potential V0 is used to measure the kinetic energy of ejected electrons. The stopping potential is the voltage required to stop the most energetic electrons from reaching the detector. The kinetic energy can be related to the stopping potential by:

K.E.=eV0

where e is the electron charge (1.602x10−19 C).

  1. Energy Conservation: The total energy of the incident photon is conserved, with part of it used to overcome the work function and the rest becoming the kinetic energy of the electron:

hf=ϕ+K.E 

There are three things that affect the photoelectric effect:

1. The intensity of the light on the cathode affects the current but does not change the energy the electrons have.


2. The frequency of the light on the cathode affects the energy of the electron


3. The metal used as the cathode will change how much energy is required to produce a current




The simulation below shows these effects. Note that when the electron energy is 0.00 that's when the electron is basically detached from the metal but it isn't moving yet.

SOURCE: Mr. A. and Tony Mangiacapre, URL Link



This experiment shows 2 things:

1. The photocurrent (current due to the photons)

2. The maximum kinetic energy of the electrons which is calculated through


KEmax=hf−ϕKE_{max} = hf - \phi

where h is 6.626×10−34Jsor4.14×1015eVs6.626 \times 10^{-34} Js or 4.14 \times 10^{15} eVs

 ϕ is the work function. This is the minimum amount of energy required to remove an electron off the cathode.



The two graphs for the photoelectric effect are:

Voltage vs Photocurrent Graph.


Frequency vs KE Graph.

Applications

  1. Photovoltaic Cells: The photoelectric effect is the principle behind photovoltaic cells, which convert light energy into electrical energy.

  2. Photoelectric Sensors: Used in various applications, such as automatic lighting, safety systems, and light meters in cameras.

  3. Electron Microscopy: Helps in the development of techniques to study materials at an atomic scale.

Quantum Implications

The photoelectric effect was pivotal in establishing the quantum theory of light. It demonstrated that light behaves not only as a wave but also as particles (photons), leading to the dual nature of light. This duality is fundamental to quantum mechanics, influencing the understanding of various phenomena at the atomic and subatomic levels.

Conclusion

The photoelectric effect is a critical phenomenon in the history of physics, bridging classical and quantum theories. It provided compelling evidence for the quantum nature of light and laid the groundwork for quantum mechanics, revolutionizing our understanding of the microscopic world.



Example 1
Light with a wavelength of 400 nm400 \, nm ejects electrons with a kinetic energy of 2.1 eV2.1 \, eV. Calculate the work function of the material.

 Use hf=ϕ+K.E.hf = \phi + K.E. and E=hcλE = \frac{hc}{\lambda}. ϕ=hcλ−K.E.\phi = \frac{hc}{\lambda} - K.E.


ϕ=400×10−9m(6.626×10−34Js)(3×108m/s)​−2.1eV

ϕ=3.1eV−2.1eV=1.0eV

ϕ=3.1 eV−2.1 eV=1.0

Example 2
The work function of a material is 2.0 eV2.0 \, eV. Calculate its threshold frequency.

 Use ϕ=hf0\phi = hf_0.

f0=ϕh=2.0×1.602×10−19 J6.626×10−34 Jsf_0 = \frac{\phi}{h} = \frac{2.0 \times 1.602 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{Js}} f0=4.83×1014 Hzf_0 = 4.83 \times 10^{14} \, \text{Hz}


Example 3
Light with a frequency of 6×1014 Hz6 \times 10^{14} \, Hz causes electrons to be emitted with a stopping potential of 1.5 V1.5 \, V. Calculate the work function.

Use eV0=hf−ϕeV_0 = hf - \phi. ϕ=hf−eV0\phi = hf - eV_0

ϕ=(6.626×10−34 Js)(6×1014 Hz)−(1.602×10−19 C)(1.5 V)\phi = (6.626 \times 10^{-34} \, \text{Js})(6 \times 10^{14} \, \text{Hz}) - (1.602 \times 10^{-19} \, \text{C})(1.5 \, V)

ϕ=3.98×10−19 J−2.403×10−19 J=1.577×10−19 J=0.98 eV\phi = 3.98 \times 10^{-19} \, \text{J} - 2.403 \times 10^{-19} \, \text{J} = 1.577 \times 10^{-19} \, \text{J} = 0.98 \, eV


Example 4
 Light of wavelength 300 nm300 \, nm strikes a material with a work function of 2.5 eV2.5 \, eV. Calculate the maximum kinetic energy of the ejected electrons.

Use K.E.=hf−ϕK.E. = hf - \phi and E=hcλE = \frac{hc}{\lambda}.

K.E.=(6.626×10−34 Js)(3×108 m/s)300×10−9 m−2.5 eVK.E. = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{300 \times 10^{-9} \, \text{m}} - 2.5 \, eV

K.E.=4.14 eV−2.5 eV=1.64 eVK.E. = 4.14 \, eV - 2.5 \, eV = 1.64 \, eV


Example 5
 What is the wavelength of light that just ejects electrons (zero kinetic energy) from a material with a work function of 3.0 eV3.0 \, eV?

Use ϕ=hcλ\phi = \frac{hc}{\lambda}. λ=hcϕ=(6.626×10−34 Js)(3×108 m/s)3×1.602×10−19 J\lambda = \frac{hc}{\phi} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{3 \times 1.602 \times 10^{-19} \, \text{J}}

λ=414 nm\lambda = 414 \, nm


Example 6
 Calculate the energy of a photon with a wavelength of 500 nm500 \, nm.

Use E=hcλE = \frac{hc}{\lambda}. E=(6.626×10−34 Js)(3×108 m/s)500×10−9 mE = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{500 \times 10^{-9} \, \text{m}}

E=3.97×10−19 J=2.48 eVE = 3.97 \times 10^{-19} \, \text{J} = 2.48 \, eV


Example 7
What is the frequency of light with a wavelength of 450 nm450 \, nm?

Use f=cλf = \frac{c}{\lambda}.

f=3×108 m/s450×10−9 mf = \frac{3 \times 10^8 \, \text{m/s}}{450 \times 10^{-9} \, \text{m}}

f=6.67×1014Hz

Exercise &&1&& (&&1&& Question)

 Light with an intensity of 10 W/m210 \, W/m^2 and wavelength 400 nm400 \, nm falls on a material. Calculate the number of electrons ejected per second per square meter if the work function is 2.2 eV2.2 \, eV.

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Exercise &&2&& (&&1&& Question)

Light of wavelength 300 nm300 \, nm falls on a material with a work function of 2.0 eV2.0 \, eV. If the light intensity is 20 W/m220 \, W/m^2, calculate the current produced assuming 100% efficiency.

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Exercise &&3&& (&&1&& Question)

How does the intensity of light affect the number of ejected electrons?

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Exercise &&4&& (&&1&& Question)

Light of wavelength 450 nm450 \, nm ejects electrons from a metal with a work function of 2.5 eV2.5 \, eV. What is the maximum kinetic energy of the ejected electrons?

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