According to Einstein’s photoelectric equation, the plot of the Kinetic Energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope:
Is the same for all metals and independent of the intensity of the radiation
Understanding Einstein's Photoelectric Equation
The question asks about the plot of the kinetic energy of emitted photoelectrons versus the frequency of the incident radiation based on Einstein's photoelectric equation. Let's first understand this fundamental equation.
Einstein's photoelectric equation is given by:
\begin{equation*} KE_{max} = h\nu - \phi_0 \end{equation*}
Where:
- \(KE_{max}\) is the maximum kinetic energy of the emitted photoelectrons.
- \(h\) is Planck's constant, a fundamental universal constant.
- \(\nu\) is the frequency of the incident radiation.
- \(\phi_0\) is the work function of the metal, which is the minimum energy required to remove an electron from the surface of that specific metal.
This equation describes the energy transfer during the photoelectric effect. When a photon of energy \(h\nu\) strikes a metal surface, if its energy is greater than or equal to the work function \(\phi_0\), an electron can be emitted. The excess energy (\(h\nu - \phi_0\)) is converted into the kinetic energy of the emitted electron, with \(KE_{max}\) being the maximum possible kinetic energy.
Plotting Kinetic Energy vs Frequency
The equation \(KE_{max} = h\nu - \phi_0\) can be rearranged into the form of a linear equation \(y = mx + c\), where \(y = KE_{max}\) and \(x = \nu\). Comparing the equation to \(y = mx + c\):
- \(y\) corresponds to \(KE_{max}\) (Kinetic Energy)
- \(x\) corresponds to \(\nu\) (Frequency)
- \(m\) corresponds to \(h\) (Slope)
- \(c\) corresponds to \(-\phi_0\) (Y-intercept)
Therefore, plotting the maximum kinetic energy (\(KE_{max}\)) of the emitted photoelectrons on the y-axis against the frequency (\(\nu\)) of the incident radiation on the x-axis will result in a straight line. The slope of this straight line is equal to Planck's constant, \(h\).
Analyzing the Slope of the KE vs Frequency Plot
The slope of the plot of \(KE_{max}\) versus \(\nu\) is equal to Planck's constant \(h\). Let's consider what Planck's constant depends on:
- Planck's constant (\(h\)) is a fundamental physical constant of nature. It has a fixed value, approximately \(6.626 \times 10^{-34}\) joule-seconds.
- It does not depend on the material of the metal surface.
- It does not depend on the intensity of the incident radiation.
- It does not depend on the frequency of the incident radiation.
Thus, the slope of the \(KE_{max}\) vs \(\nu\) plot is always equal to \(h\), regardless of the metal used or the intensity of the incident radiation.
Evaluating the Options
Let's examine the given options based on our understanding:
Option 1: Depends on the nature of the metal used
- The nature of the metal affects the work function (\(\phi_0\)), which is the y-intercept of the plot (\(-\phi_0\)). However, the slope is \(h\), which is independent of the metal's nature. This option is incorrect.
Option 2: Depends on the intensity of the radiation
- The intensity of the radiation affects the number of photons incident per unit area per unit time, and consequently, the number of photoelectrons emitted (photocurrent), provided the frequency is above the threshold frequency. However, the energy of individual photons (\(h\nu\)) and thus the maximum kinetic energy of individual photoelectrons (\(h\nu - \phi_0\)) does not depend on intensity. The slope \(h\) is also independent of intensity. This option is incorrect.
Option 3: Depends both on the intensity of the radiation and the metal used
- As explained above, the slope \(h\) depends on neither the intensity of the radiation nor the nature of the metal. This option is incorrect.
Option 4: Is the same for all metals and independent of the intensity of the radiation
- The slope of the plot is \(h\). Planck's constant \(h\) is a universal constant, meaning it is the same for all materials (metals) and does not depend on factors like the intensity of the incident radiation. This option accurately describes the slope.
Therefore, the plot of the Kinetic Energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope is the same for all metals and independent of the intensity of the radiation.
| Parameter in Plot \(y = mx + c\) | Corresponds to (Photoelectric Effect) | Description | Dependence |
|---|---|---|---|
| \(y\) | \(KE_{max}\) | Maximum Kinetic Energy of Photoelectron | Depends on \(\nu\) and \(\phi_0\) |
| \(x\) | \(\nu\) | Frequency of Incident Radiation | Independent variable |
| \(m\) (Slope) | \(h\) | Planck's Constant | Universal Constant (Independent of metal, intensity, frequency) |
| \(c\) (Y-intercept) | \(-\phi_0\) | Negative of Work Function | Depends on the nature of the metal |
Photoelectric Effect Revision Table
| Concept | Description | Related to |
|---|---|---|
| Photoelectric Effect | Emission of electrons from a metal surface when light of suitable frequency falls on it. | Light as particles (photons) |
| Photon | A quantum of light energy, with energy \(E = h\nu\). | Planck's constant (\(h\)) and frequency (\(\nu\)) |
| Work Function (\(\phi_0\)) | Minimum energy required to remove an electron from a specific metal surface. | Nature of the metal |
| Threshold Frequency (\(\nu_0\)) | Minimum frequency of incident light required for photoelectric emission (\(h\nu_0 = \phi_0\)). | Work function (\(\phi_0\)) and nature of the metal |
| Stopping Potential (\(V_s\)) | Minimum negative potential applied to the collector plate to stop the most energetic photoelectrons. \(eV_s = KE_{max}\). | Maximum kinetic energy (\(KE_{max}\)) |
| Photocurrent | The flow of emitted photoelectrons. | Intensity of incident radiation (for \(\nu > \nu_0\)) |
Additional Information on Photoelectric Effect
Beyond the slope of the KE vs frequency plot, there are other key aspects of the photoelectric effect:
- Threshold Frequency: For electron emission to occur, the frequency of the incident light (\(\nu\)) must be greater than or equal to a minimum frequency called the threshold frequency (\(\nu_0\)). If \(\nu < \nu_0\), no electrons are emitted, regardless of how high the intensity is. The threshold frequency is related to the work function by \(\phi_0 = h\nu_0\).
- Work Function: Different metals have different work functions. A metal with a lower work function will have a lower threshold frequency, meaning it requires less energetic photons to cause electron emission.
- Intensity vs Photocurrent: For a given frequency above the threshold frequency, increasing the intensity of the incident light increases the number of photons striking the metal surface per second. Each photon (with sufficient energy) can eject one electron. Therefore, increasing the intensity increases the number of emitted electrons, resulting in a higher photocurrent. However, the maximum kinetic energy of each emitted electron remains unchanged, as it depends only on the frequency of the light and the work function of the metal (\(KE_{max} = h\nu - \phi_0\)).
- Instantaneous Process: The photoelectric effect is essentially an instantaneous process. Electron emission occurs almost immediately upon the incidence of light, provided the frequency is above the threshold, showing that the energy is transferred in discrete packets (photons). This was a key piece of evidence supporting the particle nature of light.
Important Questions from Dual Nature of Radiation and Matter
The work function for an Aluminium surface is 4.2 eV. Find the threshold wavelength for the photoelectric emission.
A potentiometer wire of length L and a resistance r are connected in series with a battery of emf E0 and a resistance r1. An unknown emf E is balanced at a length l of the potentiometer wire. The emf E will be:
The time taken by light to travel normally through a glass plate of thickness 1 mm would be:
(Take refractive index of glass = 1.5)
Energy of a photon corresponding to a wavelength of 600 nm is 2.08 eV. The energy of a photon of wavelength 400 nm will be:
A particle moves three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10-4. The mass of the particle is:
