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Modern Physics
Photoelectric Effect
The photoelectric effect is the emission of electrons from a metallic surface when electromagnetic radiation of sufficiently high frequency falls upon it. It provides definitive proof of the particle nature of light through Einstein's photon hypothesis.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Basic definition, threshold frequency, work function, Einstein's photoelectric equation, and dependence of stopping potential on frequency and intensity.
PGT extension
Detailed analysis of stopping potential vs. frequency graphs, determination of Planck's constant, particle nature of photons, momentum of photons, and relativistic photoelectric considerations.
Prerequisites
Wave-particle duality concepts
Conservation of energy
Work function and binding energy of electrons in metals
Basic Phenomenon and Experimental Observations
Photoelectric emission is an instantaneous process with no measurable time lag between light incidence and electron ejection.
Emission occurs only if the incident frequency exceeds a specific minimum frequency called the threshold frequency (ν0).
The maximum kinetic energy of emitted photoelectrons is independent of the intensity of incident light and depends solely on its frequency.
Einstein's Explanation and Photon Hypothesis
Light propagates in discrete packets of energy called photons, each having energy $E = h\nu$.
A single photon transfers its entire energy to a single electron in an elastic-like collision at the surface.
The maximum kinetic energy of the ejected electron is governed by the conservation of energy: Kmax = hν - φ, where φ is the work function.
Role of Intensity and Frequency
Increasing the intensity of incident light (at a constant frequency) increases the number of incident photons per second, thereby increasing the photoelectric current.
Increasing the frequency of incident light increases the energy of individual photons, resulting in higher maximum kinetic energy and greater stopping potential.
Below the threshold frequency, no photoelectrons are emitted regardless of how high the incident light intensity is.
Stopping Potential Concept
Stopping potential (Vs) is the minimum negative potential applied to the anode that reduces the photoelectric current to zero.
It provides a direct measure of the maximum kinetic energy of the photoelectrons via the relation e Vs = Kmax.
The slope of the stopping potential versus frequency graph gives the universal constant ratio h/e.
Derivations
Derivation of Einstein's Photoelectric Equation and Stopping Potential
Conditions: Monochromatic light of frequency ν incident on a clean metal surface with work function φ
Apply conservation of energy to the interaction between a single photon and an electron: Energy of photo$n = Work function + Maximum kinetic energy of emitted electron$.
Substitute the expressions: h$\nu = \phi + K_{max}$, where $\phi = h\nu_{0} (\nu_{0}$ is threshold frequency).
Express maximum kinetic energy in terms of stopping potential Vs: $K_{max} = e V_{s}$.
Combine the relations to obtain the final voltage-frequency equation: e $V_{s} = h\nu - h\nu_{0}$.
Result: $V_{s} = \frac{h}{e}(\nu - \nu_{0})$
Derivation of Photon Momentum and Radiation Pressure
Conditions: Photons considered as relativistic particles with zero rest mass
Use the energy-momentum relation for a particle with zero rest mass: E = pc, where p is momentum and c is the speed of light.
Substitute the photon energy E = h\nu: pc = h\nu.
Express momentum in terms of wavelength λ using $c = \nu\lambda: p = \frac{h\nu}{c} = \frac{h}{\lambda}$.
For complete absorption at a surface, force F = \frac{dn}{dt} p, where \frac{dn}{dt} is the number of photons striking per second.
Result: p = \frac{h}{\lambda}
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Energy of a Photon
$$E = h\nu = \frac{hc}{\lambda}$$
h = Planck's constant, $\nu = frequency$, c = speed of light, $\lambda = wavelength$ Joules (J)
Valid for all electromagnetic radiation
Einstein's Photoelectric Equation
$$K_{max} = h\nu - \phi$$
$K_{max} = max kinetic energy$, h$\nu = photon energy$, $\phi = work function$ Joules (J) or electron-volts (eV)
Icurrent = photoelectric current, Nphotons = number of incident photons per second Amperes (A)
At constant frequency above threshold
Graphs and experiments
Stopping Potential vs. Frequency Graph
Draw: A straight line plot of stopping potential (Vs) on the y-axis against frequency (ν) on the x-axis, intersecting the x-axis at ν0.
Interpret: The intercept on the frequency axis gives the threshold frequency ν0. The slope of the line is equal to h/e, independent of the metal used.
Exam use: Frequently used to calculate Planck's constant experimentally and to determine work function.
Photoelectric Current vs. Anode Potential
Draw: Curves showing saturation current at positive potentials and dropping to zero at a negative stopping potential (-Vs) for different light intensities at constant frequency.
Interpret: Saturation current is directly proportional to intensity, but the stopping potential remains identical for all curves sharing the same frequency.
Exam use: Tests understanding of the independence of maximum kinetic energy from light intensity.
Lenard's Experimental Setup
Draw: An evacuated quartz bulb containing an emitter plate (C) and a collector plate (A), connected to a variable voltage source, reversing switch, and sensitive microammeter.
Interpret: Demonstrates the control of electron flow using retarding and accelerating potentials under monochromatic illumination.
Exam use: Conceptual questions on experimental design, vacuum requirements, and role of quartz windows.
Solved numericals
TGT core
The work function of a metal is 2.0 eV. Calculate its threshold wavelength. (Take hc = 12400 eV Å)
Identify given values: $\phi = 2.0 eV$, hc = 12400 eV Å.
Use formula: \lambd$a_{0} = \frac{hc}{\phi}$.
Substitute values: \lambd$a_{0} = \frac{12400\text{ eV Å}}{2.0\text{ eV}}$.
Calculate final value: \lambd$a_{0} = 6200\text{ Å}$.
Answer: 6200 Å
PGT extension
Light of wavelength 4000 Å falls on a photosensitive metal with work function 2.5 eV. Find the maximum velocity of the emitted photoelectrons. (Mass of electron $m = 9.1 \times 10^{-31}\text{ kg}$, $h = 6.63 \times 10^{-34}\text{ J s}$, $c = 3 \times 10^8\text{ m s}^{-1}$, 1\text{ eV} = 1.6 \times 10^{-19}\text{ J})
Calculate energy of incident photon in Joules: $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10}} = 4.9725 \times 10^{-19}\text{ J}$.
Convert work function to Joules: $\phi = 2.5 \times 1.6 \times 10^{-19} = 4.0 \times 10^{-19}\text{ J}$.
When monochromatic light of frequency 1.5 \times 10^{15}\text{ Hz} is incident on a metallic plate, the stopping potential is found to be 2.0 V. When the frequency is changed to 2.0 \times 10^{15}\text{ Hz}, the stopping potential becomes 4.0 V. Calculate the value of Planck's constant from these data, given the electronic charge $e = 1.6 \times 10^{-19}\text{ C}$.
Write the stopping potential equations for both frequencies: e $V_{s1} = h\nu_{1} - \phi$ and e $V_{s2} = h\nu_{2} - \phi$.
Subtract the first equation from the second: e(V_{s2} - V_{s1}) = h(\nu_{2} - \nu_{1}).
Substitute the given values: 1.6 \times 10^{-19}(4.0 - 2.0) = h(2.0 \times 10^{15} - 1.5 \times 10^{15}).
Assumption: Single-photon absorption by a single electron without multiple scattering events inside the metal. Limitation: High-intensity laser beams can induce multi-photon photoelectric emission.
Assumption: Zero rest mass approximation for photons interacting with non-relativistic electrons. Limitation: At extremely high gamma-ray energies, pair production competes with photoelectric emission.
Common misconceptions
Common error: Increasing the intensity of incident light increases the kinetic energy of emitted photoelectrons.
Correct: Increasing light intensity increases the number of emitted photoelectrons (current), but their maximum kinetic energy depends only on frequency.
Intensity governs the photon flux density, whereas photon frequency determines individual photon energy.
Common error: Photoelectric emission takes time to accumulate sufficient energy if the light is very dim.
Correct: Photoelectric emission is instantaneous once the threshold frequency condition is met.
Energy transfer happens via localized photon packets instantaneously, not via continuous classical wave energy accumulation.
Common error: Any frequency of light can cause photoelectric emission if the exposure time is long enough.
Correct: Light with frequency below the threshold frequency produces no photoelectrons, regardless of duration or intensity.
A single photon must carry energy greater than or equal to the work function to eject an electron.
Quick revision
Photoelectric emission proves the quantum particle nature of light.
Threshold frequency is the minimum frequency required for electron emission.
Stopping potential measures the maximum kinetic energy of photoelectrons.
Einstein's photoelectric equation is based on conservation of energy in photon-electron collisions.
Must remember
Photoelectric emission is an instantaneous process with no measurable time lag.
The maximum kinetic energy ($K_{max}$) of emitted photoelectrons is independent of the intensity of incident light and depends solely on its frequency.
Einstein's Photoelectric Equation is given by $K_{max} = h\nu - \phi$.
Stopping potential ($V_s$) provides a direct measure of the maximum kinetic energy via the relation $e V_s = K_{max}$.
The slope of the stopping potential versus frequency graph yields the universal constant ratio $h/e$.
Below the threshold frequency ($\nu_0$), no photoelectrons are emitted regardless of how high the incident light intensity is.
Increasing the intensity of incident light at a constant frequency increases the number of incident photons per second, thereby increasing the photoelectric current.
Glossary
Threshold Frequency ($\nu_0$)The minimum frequency of incident radiation required to eject photoelectrons from a metal surface.
Work Function ($\phi$)The minimum energy required to liberate an electron from the surface of a metal.
Stopping Potential ($V_s$)The minimum negative potential applied to the anode that reduces the photoelectric current to zero.
PhotonA discrete packet or quantum of electromagnetic energy having energy $E = h\nu$.
Effect of Intensity vs. Frequency on Photoelectric Emission
Parameter Changed
Effect on Photoelectric Current
Effect on Maximum Kinetic Energy
Increasing Intensity (Constant Frequency)
Increases
Remains Unchanged
Increasing Frequency (Constant Intensity)
Remains Unchanged
Increases
Practice Quiz
Choose an option to reveal the explanation.
1. What is the nature of the time lag between the incidence of light and the ejection of photoelectrons?
Explanation: Photoelectric emission is an instantaneous process with no measurable time lag because energy transfer occurs via single-photon single-electron localized packets.
2. Which of the following determines the maximum kinetic energy of emitted photoelectrons?
Explanation: According to Einstein's photoelectric equation $K_{max} = h\nu - \phi$, the maximum kinetic energy depends solely on the frequency of the incident radiation.
3. Below which characteristic parameter will no photoelectrons be emitted from a metal surface, no matter how high the intensity is?
Explanation: If the incident frequency is below the threshold frequency ($\nu_0$), no photoelectrons are emitted regardless of the intensity of the light.
4. What is represented by the slope of the stopping potential ($V_s$) versus frequency ($\nu$) graph?
Explanation: From the relation $eV_s = h\nu - \phi$, rearranging gives $V_s = (h/e)\nu - \phi/e$. Thus, the slope of $V_s$ versus $\nu$ is $h/e$.
5. If the frequency of incident light on a metal surface is doubled, what happens to the maximum kinetic energy of the emitted photoelectrons?
Explanation: Since $K_{max} = h\nu - \phi$, doubling $\nu$ makes the new kinetic energy $2h\nu - \phi$, which is clearly more than double the initial kinetic energy ($h\nu - \phi$).
6. What is the momentum of a photon of frequency $\nu$ and wavelength $\lambda$?
Explanation: The momentum of a photon is given by the de Broglie and photon relation $p = \frac{h}{\lambda} = \frac{E}{c}$.
7. How does increasing the intensity of incident light (at a frequency above threshold) affect the photoelectric current?
Explanation: Increasing intensity increases the number of incident photons per second, which proportionally increases the number of emitted photoelectrons and hence the photoelectric current.
8. What is the expression for the threshold frequency $\nu_0$ in terms of the work function $\phi$?
Explanation: Since $\phi = h\nu_0$, the threshold frequency is given by $\nu_0 = \frac{\phi}{h}$.
9. Which physical quantity does the stopping potential ($V_s$) directly measure?
Explanation: Stopping potential provides a direct measure of the maximum kinetic energy via the relation $eV_s = K_{max}$.
10. What is the rest mass of a photon?
Explanation: Photons are massless relativistic particles and have a rest mass of zero.
11. If the wavelength of incident light is increased beyond the threshold wavelength $\lambda_0$, what happens to photoelectron emission?
Explanation: A wavelength greater than $\lambda_0$ corresponds to a frequency less than $\nu_0$, which results in no photoelectric emission.
12. What are the units of Planck's constant $h$?
Explanation: Since $E = h\nu$, the dimensions of $h$ are Energy $\times$ Time, yielding Joule second ($\text{J s}$).
13. Which of the following correctly gives the maximum velocity ($v_{max}$) of photoelectrons?
Explanation: From $K_{max} = \frac{1}{2}mv_{max}^2 = h\nu - \phi$, we get $v_{max} = \sqrt{\frac{2(h\nu - \phi)}{m}}$.
14. In a photoelectric setup, saturation current depends upon:
Explanation: Saturation current is directly proportional to the intensity of incident light because higher intensity means more photons striking per second and thus more electrons emitted.
15. Why is classical wave theory inadequate to explain the photoelectric effect?
Explanation: Classical wave theory suggests energy is spread out over wavefronts, requiring a time lag for dim light to accumulate enough energy. Experimentally, emission is instantaneous.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. Light of frequency $1.5\nu_0$ is incident on a photosensitive material, where $\nu_0$ is the threshold frequency. If the frequency of incident light is halved and the intensity is doubled, what will be the photoelectric current?
Explanation: The initial frequency is $1.5\nu_0$. When halved, the new frequency becomes $0.75\nu_0$, which is below the threshold frequency $\nu_0$. Therefore, no photoelectrons are emitted, and the photoelectric current becomes zero.
2. The work function of a metal is $2.0\,\text{eV}$. Light of wavelength $300\,\text{nm}$ is incident on it. What is the maximum kinetic energy of the emitted photoelectrons? (Take $hc \approx 1240\,\text{eV nm}$)
Explanation: Photon energy $E = \frac{hc}{\lambda} = \frac{1240\,\text{eV nm}}{300\,\text{nm}} \approx 4.13\,\text{eV}$. Maximum kinetic energy $K_{max} = E - \phi = 4.13\,\text{eV} - 2.0\,\text{eV} = 2.13\,\text{eV}$.
3. Stopping potential for a certain photosensitive metal is $V_1$ when frequency $\nu_1$ is used, and $V_2$ when frequency $\nu_2$ is used. What is the value of Planck's constant $h$ in terms of $V_1, V_2, \nu_1, \nu_2, e$?
Explanation: From $eV_1 = h\nu_1 - \phi$ and $eV_2 = h\nu_2 - \phi$, subtracting the two equations yields $e(V_2 - V_1) = h(\nu_2 - \nu_1)$, leading to $h = \frac{e(V_2 - V_1)}{\nu_2 - \nu_1}$.
4. If the threshold wavelength for a metal is $5000\,\text{\AA}$, its work function is:
5. In a photoelectric experiment, graph of stopping potential $V_s$ versus frequency $\nu$ makes an intercept on the frequency axis equal to:
Explanation: From $V_s = \frac{h}{e}\nu - \frac{\phi}{e}$, when $V_s = 0$, $\nu = \frac{\phi}{h} = \nu_0$. Thus, the intercept on the frequency axis represents the threshold frequency.
6. Monochromatic light of frequency $\nu$ is incident on a photocell. The stopping potential is found to be $V_s$. If the frequency of light is made $2\nu$, the new stopping potential will be:
Explanation: Initial relation: $eV_s = h\nu - \phi$. New relation: $eV_s' = h(2\nu) - \phi = 2h\nu - \phi$. Notice that $2h\nu - \phi = 2(h\nu - \phi) + \phi = 2eV_s + \phi$, which means $V_s' = 2V_s + \frac{\phi}{e}$. Thus, $V_s'$ is strictly greater than $2V_s$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What happens to the maximum velocity of photoelectrons if the intensity of incident light is increased while keeping frequency constant?
Explanation: Intensity affects the number of photons and hence current, but individual photon energy and maximum kinetic energy/velocity depend only on frequency.
2. The work function for a metal is $4.14\,\text{eV}$. The threshold frequency for this metal is approximately:
5. A photoelectric surface is illuminated by successively different monochromatic lights. The maximum kinetic energy of electrons is plotted against frequency. The intercept of this line on the frequency axis gives:
Explanation: In the $K_{max}$ versus $\nu$ graph, $K_{max} = h\nu - h\nu_0$. When $K_{max} = 0$, $\nu = \nu_0$, which is the intercept on the frequency axis.
6. If the wavelength of incident light is halved, how does the maximum kinetic energy of photoelectrons change, assuming initial frequency was well above threshold?
Explanation: Halving wavelength doubles frequency $\nu$. Since $K_{max} = h\nu - \phi$, doubling $\nu$ increases $K_{max}$ to more than double the original value.
7. Photoelectric current is independent of which of the following once saturation is reached?
Explanation: Once saturation current is reached, further increases in positive anode potential do not increase the current because all emitted photoelectrons are already collected.
8. The ratio of the de Broglie wavelength of an electron to that of a photon having the same energy $E$ is proportional to:
Explanation: For an electron, $\lambda_e = \frac{h}{\sqrt{2mE}}$. For a photon, $\lambda_p = \frac{hc}{E}$. The ratio $\frac{\lambda_e}{\lambda_p} = \frac{E}{c\sqrt{2mE}} = \frac{\sqrt{E}}{c\sqrt{2m}}$, which is proportional to $E^{1/2}$. Wait, let's re-verify: $\lambda_e / \lambda_p \propto E^{1/2}$.
9. Which metal generally exhibits photoelectric emission even with visible light?
Explanation: Cesium has a very low work function, allowing it to exhibit photoelectric emission under visible light.
10. In the context of the particle nature of light, what does Einstein's photoelectric equation validate?
Explanation: Einstein's photoelectric equation successfully validates the quantum/particle hypothesis where light interacts in discrete energy packets (photons).