Photon energy \[6eV\] is incident on a metal surface of work function \[4\,eV\]. Maximum KE of emitted photo-electrons will be:
A. \[0\,eV\]
B. \[1\,eV\]
C. \[2\,eV\]
D. \[10\,eV\]
Answer
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Hint: Here the concept of photoelectric effect will be applied to solve the problem. According to this the total energy of a photon is equal to the sum of the energy utilized to eject an electron and the maximum kinetic energy of electrons. By using this we can obtain the maximum KE of emitted photo-electrons as the maximum kinetic energy of electrons is equal to the energy of the incident light energy packet minus the work function.
Formula used:
Kinetic energy (\[K{E_{\max }}\]) is given as:
\[K{E_{\max }} = E - \phi \]
Where E is the energy of the incident radiation and \[\phi \] is the work function.
Complete step by step solution:
Given Energy of photon, \[E = 6\,eV\]
Work function, \[\phi = 4\,eV\]
As we know that,
\[K{E_{\max }} = E - \phi \]
By substituting the values, we get
\[K{E_{\max }}= 6\,eV - 4\,eV\]
\[\therefore K{E_{\max }}= 2\,eV\]
Therefore, the maximum KE of emitted photo-electrons will be \[2\,eV\].
Hence option C is the correct answer.
Additional information: Photon energy is the energy carried by a single photon. The energy is directly related to the photon's electromagnetic frequency and is inversely related to the wavelength. The higher the photon's frequency the energy will be higher. The energy needed by the particle to come from the medium and break to the surface. Photoelectric is the phenomenon where electrons are ejected from a metal surface when the light of sufficient frequency is incident on it. Einstein suggested that light behaved like a particle and that each particle of light has energy called a photon.
Note: Students routinely make mistakes while writing Einstein's photoelectric effect equation. Remember that the energy of incoming radiation is computed as the sum of the photoelectron's kinetic energy and the metal's work function.
Formula used:
Kinetic energy (\[K{E_{\max }}\]) is given as:
\[K{E_{\max }} = E - \phi \]
Where E is the energy of the incident radiation and \[\phi \] is the work function.
Complete step by step solution:
Given Energy of photon, \[E = 6\,eV\]
Work function, \[\phi = 4\,eV\]
As we know that,
\[K{E_{\max }} = E - \phi \]
By substituting the values, we get
\[K{E_{\max }}= 6\,eV - 4\,eV\]
\[\therefore K{E_{\max }}= 2\,eV\]
Therefore, the maximum KE of emitted photo-electrons will be \[2\,eV\].
Hence option C is the correct answer.
Additional information: Photon energy is the energy carried by a single photon. The energy is directly related to the photon's electromagnetic frequency and is inversely related to the wavelength. The higher the photon's frequency the energy will be higher. The energy needed by the particle to come from the medium and break to the surface. Photoelectric is the phenomenon where electrons are ejected from a metal surface when the light of sufficient frequency is incident on it. Einstein suggested that light behaved like a particle and that each particle of light has energy called a photon.
Note: Students routinely make mistakes while writing Einstein's photoelectric effect equation. Remember that the energy of incoming radiation is computed as the sum of the photoelectron's kinetic energy and the metal's work function.
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