Radiation of frequency 2ν0 is incident on a metal with threshold frequency ν0. The correct statement is:
Maximum kinetic energy of photoelectrons emitted can be hν0
Understanding the Photoelectric Effect: Frequency and Kinetic Energy
The question asks us to determine the correct statement about photoelectrons emitted when radiation of a specific frequency is incident on a metal with a given threshold frequency. This involves applying the principles of the photoelectric effect.
The photoelectric effect is a phenomenon where electrons are emitted from a metal surface when light of sufficient frequency is incident upon it. The minimum frequency required to cause emission is called the threshold frequency (\(\nu_0\)). The minimum energy required to remove an electron from the metal surface is called the work function (\(\phi\)). The work function is related to the threshold frequency by the equation:
\(\phi = h\nu_0\)
where \(h\) is Planck's constant.
According to Einstein's photoelectric equation, the energy of the incident photon (\(h\nu\)) is used to overcome the work function (\(\phi\)) and provide kinetic energy (\(K\)) to the emitted electron. The maximum kinetic energy (\(K_{max}\)) of the emitted photoelectron is given by:
\(h\nu = \phi + K_{max}\)
Rearranging the equation to find the maximum kinetic energy:
\(K_{max} = h\nu - \phi\)
Applying the Given Values to the Photoelectric Effect
In this problem, we are given:
- Incident radiation frequency (\(\nu\)) = \(2\nu_0\)
- Threshold frequency of the metal (\(\nu_0\))
First, let's confirm if photoemission will occur. Photoemission happens if the incident frequency (\(\nu\)) is greater than or equal to the threshold frequency (\(\nu_0\)). Here, \(2\nu_0 > \nu_0\) (assuming \(\nu_0 > 0\)), so photoelectrons will be emitted.
Next, let's calculate the maximum kinetic energy of the emitted photoelectrons using Einstein's equation:
\(K_{max} = h\nu - \phi\)
We know that \(\nu = 2\nu_0\) and \(\phi = h\nu_0\). Substitute these values into the equation:
\(K_{max} = h(2\nu_0) - h\nu_0\)
\(K_{max} = 2h\nu_0 - h\nu_0\)
\(K_{max} = h\nu_0\)
So, the maximum kinetic energy of the photoelectrons emitted will be \(h\nu_0\).
Analyzing the Given Options
Let's examine each option in light of our understanding and calculation:
- No photoelectrons will be emitted
- All photoelectrons emitted will have kinetic energy equal to \(h\nu_0\)
- Maximum kinetic energy of photoelectrons emitted can be \(h\nu_0\)
- Maximum kinetic energy of photoelectrons emitted will be \(2h\nu_0\)
Analysis:
- Option 1: This is incorrect because the incident frequency (\(2\nu_0\)) is greater than the threshold frequency (\(\nu_0\)), so photoemission will occur.
- Option 2: This is incorrect. \(h\nu_0\) is the *maximum* kinetic energy. Electrons emitted from the surface have the maximum kinetic energy, but electrons from deeper inside the metal lose some energy due to collisions before escaping, so their kinetic energy will be less than the maximum. Therefore, not *all* emitted photoelectrons will have this kinetic energy.
- Option 3: This states that the maximum kinetic energy can be \(h\nu_0\). Our calculation showed that the maximum kinetic energy \(K_{max}\) is exactly \(h\nu_0\). While the phrasing "can be" is slightly imprecise (it *will be* \(h\nu_0\) for the most energetic electrons), among the given options, this is the statement that correctly identifies the value of the maximum kinetic energy. The actual kinetic energy of individual photoelectrons ranges from 0 up to this maximum value.
- Option 4: This is incorrect. Our calculation showed that the maximum kinetic energy is \(h\nu_0\), not \(2h\nu_0\).
Based on the calculation and analysis, option 3 correctly states the maximum possible kinetic energy of the emitted photoelectrons.
Conclusion on Photoelectron Kinetic Energy
When radiation of frequency \(2\nu_0\) is incident on a metal with threshold frequency \(\nu_0\), photoemission occurs. The maximum kinetic energy of the emitted photoelectrons is given by the energy difference between the incident photon and the work function, which we calculated to be \(h\nu_0\). Therefore, the statement that the maximum kinetic energy of photoelectrons emitted can be \(h\nu_0\) is the correct one among the given options.
| Parameter | Value | Formula/Relation |
|---|---|---|
| Incident Frequency (\(\nu\)) | \(2\nu_0\) | Given |
| Threshold Frequency (\(\nu_0\)) | \(\nu_0\) | Given |
| Work Function (\(\phi\)) | \(h\nu_0\) | \(\phi = h\nu_0\) |
| Condition for Emission | \(2\nu_0 \ge \nu_0\) | \(\nu \ge \nu_0\) |
| Maximum Kinetic Energy (\(K_{max}\)) | \(h\nu_0\) | \(K_{max} = h\nu - \phi\) |
Revision Table: Key Concepts in Photoelectric Effect
| Term | Definition | Significance |
|---|---|---|
| Threshold Frequency (\(\nu_0\)) | Minimum frequency of incident light required for photoemission. | No emission occurs below this frequency, regardless of intensity. |
| Work Function (\(\phi\)) | Minimum energy required to remove an electron from the metal surface. | \(\phi = h\nu_0\). It is a property of the metal. |
| Incident Frequency (\(\nu\)) | Frequency of the light falling on the metal. | If \(\nu \ge \nu_0\), emission occurs. Determines the energy of incident photons (\(h\nu\)). |
| Photoelectron | Electron emitted from a metal surface during the photoelectric effect. | Its kinetic energy depends on the incident photon energy and the work function. |
| Maximum Kinetic Energy (\(K_{max}\)) | Highest possible kinetic energy of an emitted photoelectron. | \(K_{max} = h\nu - \phi\). Determined by the energy difference between the photon and the work function. |
Additional Information on Photoelectric Emission
The photoelectric effect demonstrates the particle nature of light, where light energy is carried in discrete packets called photons. Each photon has energy \(E = h\nu\).
- If the photon energy \(h\nu\) is less than the work function \(\phi\), an electron cannot gain enough energy to escape, and no photoemission occurs.
- If \(h\nu \ge \phi\), the photon energy is absorbed by an electron. The electron uses energy \(\phi\) to escape the metal, and the remaining energy \(h\nu - \phi\) is converted into kinetic energy.
- Electrons that are deeper within the metal or lose energy through collisions before reaching the surface will have kinetic energy less than the maximum. The maximum kinetic energy is achieved by electrons near the surface that escape without losing energy.
- The number of emitted photoelectrons (photocurrent) is proportional to the intensity of the incident light, provided the frequency is above the threshold frequency.
- The maximum kinetic energy of the emitted photoelectrons depends only on the frequency of the incident light and the work function of the metal, not on the intensity of the light. Increasing intensity increases the number of photons, thus increasing the number of emitted electrons, but not their individual maximum energy.
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:
