
Wavelength ${\lambda _A}$ and ${\lambda _B}$ are incident on two identical metal plates and photoelectrons are emitted. If ${\lambda _A} = 2{\lambda _B}$, the maximum kinetic energy of photoelectrons is _____.
Answer
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Hint: When a quantum of light strikes a metal surface, it transfers its energy to the electrons in the metal. A part of the energy of the photon is transferred into the kinetic energy of the electron emitted.
Formula used: ${\text{K}} = \dfrac{{{\text{hc}}}}{\lambda }$
Where, ${\text{K}}$ is the maximum kinetic energy of photoelectrons.
${\text{h}}$ is the Planck’s constant.
And $\lambda $ is the wavelength of incident light
Complete step by step solution:
In order for an electron to escape from the surface of the metal, it must overcome the attractive force of the positive ions in metal. As a result, a part of the photon’s energy is absorbed by the metal surface releasing the electron. This absorbed energy is known as the work function of a metal. The remaining part of the energy of the photon is transferred into the kinetic energy of the electron.
The maximum kinetic energy that a particle can obtain will be equal to the energy of the photon that strikes the metal surface. Thus, the maximum kinetic energy can be given by ${\text{K}} = \dfrac{{{\text{hc}}}}{\lambda }$
The maximum kinetic energy for the first wavelength is ${{\text{K}}_A} = \dfrac{{{\text{hc}}}}{{{\lambda _A}}}$
It is given in the question that, ${\lambda _A} = 2{\lambda _B}$.
Thus, we get, ${{\text{K}}_A} = \dfrac{{{\text{hc}}}}{{2{\lambda _B}}}$
$ \Rightarrow {{\text{K}}_{\text{A}}} = \dfrac{{{{\text{K}}_{\text{B}}}}}{{\text{2}}}$
\[ \Rightarrow 2{{\text{K}}_{\text{A}}} = {{\text{K}}_{\text{B}}}\]
Thus, the maximum kinetic energy of the photoelectron is \[{{\text{K}}_{\text{B}}} = 2{{\text{K}}_{\text{A}}}\]
Note: In photoelectric effect, when electromagnetic radiation hits a material, electrons are emitted. Electrons emitted in this process are called photoelectrons. This effect is studied in solid state, condensed matter physics, and quantum chemistry to figure out conclusions about the properties of atoms, molecules and solids. The effect is commonly used in electronic devices specialized for light detection and precisely timed electron emission.
Formula used: ${\text{K}} = \dfrac{{{\text{hc}}}}{\lambda }$
Where, ${\text{K}}$ is the maximum kinetic energy of photoelectrons.
${\text{h}}$ is the Planck’s constant.
And $\lambda $ is the wavelength of incident light
Complete step by step solution:
In order for an electron to escape from the surface of the metal, it must overcome the attractive force of the positive ions in metal. As a result, a part of the photon’s energy is absorbed by the metal surface releasing the electron. This absorbed energy is known as the work function of a metal. The remaining part of the energy of the photon is transferred into the kinetic energy of the electron.
The maximum kinetic energy that a particle can obtain will be equal to the energy of the photon that strikes the metal surface. Thus, the maximum kinetic energy can be given by ${\text{K}} = \dfrac{{{\text{hc}}}}{\lambda }$
The maximum kinetic energy for the first wavelength is ${{\text{K}}_A} = \dfrac{{{\text{hc}}}}{{{\lambda _A}}}$
It is given in the question that, ${\lambda _A} = 2{\lambda _B}$.
Thus, we get, ${{\text{K}}_A} = \dfrac{{{\text{hc}}}}{{2{\lambda _B}}}$
$ \Rightarrow {{\text{K}}_{\text{A}}} = \dfrac{{{{\text{K}}_{\text{B}}}}}{{\text{2}}}$
\[ \Rightarrow 2{{\text{K}}_{\text{A}}} = {{\text{K}}_{\text{B}}}\]
Thus, the maximum kinetic energy of the photoelectron is \[{{\text{K}}_{\text{B}}} = 2{{\text{K}}_{\text{A}}}\]
Note: In photoelectric effect, when electromagnetic radiation hits a material, electrons are emitted. Electrons emitted in this process are called photoelectrons. This effect is studied in solid state, condensed matter physics, and quantum chemistry to figure out conclusions about the properties of atoms, molecules and solids. The effect is commonly used in electronic devices specialized for light detection and precisely timed electron emission.
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