The threshold frequency fo of Cesium is 4.83 × 10¹⁴ Hz
The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:
[tex]\large {\boxed {E = h \times f}}[/tex]
E = Energi of A Photon ( Joule )
h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )
f = Frequency of Eletromagnetic Wave ( Hz )
The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.
[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]
[tex]\large {\boxed {E = qV + \Phi}}[/tex]
E = Energi of A Photon ( Joule )
m = Mass of an Electron ( kg )
v = Electron Release Speed ( m/s )
Ф = Work Function of Metal ( Joule )
q = Charge of an Electron ( Coulomb )
V = Stopping Potential ( Volt )
Let us now tackle the problem!
Given:
Φ = 2.1 eV = 2 × 1.60 × 10⁻¹⁹ Joule = 3.20 × 10⁻¹⁹ Joule
Unknown:
fo = ?
Solution:
[tex]\Phi = h \times fo[/tex]
[tex]3.20 \times 10^{-19} = 6.63 \times 10^{-34} \times fo[/tex]
[tex]fo = ( 3.20 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )[/tex]
[tex]\large {\boxed{fo = 4.83 \times 10^{14} ~ Hz} }[/tex]
Grade: High School
Subject: Physics
Chapter: Quantum Physics
Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt
The threshold frequency of the cesium is [tex]\fbox{\begin\\5.06 \times {10^{14}}\,{\text{Hz}}\end{minispace}}[/tex].
Further Explanation:
The threshold energy is the amount of energy required by an electron so that it is ejected from the surface of the material.
Concept:
The photoelectric effect is the process in which emission of the electron from the surface of material when they acquire sufficient amount of energy from the photon (i.e. the packets of energy) to release from the surface of the material.
The threshold energy of the electron can be expressed as:
[tex]\fbox{\begin\\E = {E_k} + \phi\end{minispace}}[/tex]
Here, [tex]E[/tex] is the amount of energy supplied to the surface by a photon, [tex]{E_k}[/tex] is the amount of kinetic energy of electron after emission and [tex]\phi[/tex] is the work function of the material.
The work function of a surface is the minimum amount of energy to be incident on the surface of a metal such that it can emit the electron.
The work function of material can be expressed as:
[tex]\phi=h{\nu _0}[/tex]
Here, [tex]{\nu _0}[/tex] is the threshold frequency of the electron.
The work function of the cesium is [tex]2.1\,{\text{eV}}[/tex].
Convert the work function in [tex]eV[/tex] to Joules.
[tex]\begin{aligned}\phi&=2.1{\mkern1mu} {\text{eV}}\left({1.6 \times{{10}^{ - 19}}}\right)\,{\text{J}}\\&=3.36\times10^{-19}\text{ J}\\\end{aligned}[/tex]
Substitute [tex]3.36\times{10^{-19}}\,{\text{J}}[/tex] for [tex]\phi[/tex] and [tex]6.63\times{10^{-34}\text{ kg.m}^2/\text{s}[/tex] for [tex]h[/tex] in expression for work function.
[tex]\begin{aligned}3.36\times{10^{-19}}&=\left({6.63\times{{10}^{-34}}}\right){\nu_0}\\{\nu_0}&=\frac{{3.36\times{{10}^{-19}}}}{{6.63\times{{10}^{-34}}}}\\&=5.06\times{10^{14}}\,{\text{Hz}}\\\end{aligned}[/tex]
Thus, the threshold frequency of cesium is [tex]5.06 \times {10^{14}}\,{\text{Hz}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Photoelectric effect
Keywords:
Cesium, work function, threshold frequency, photoelectric effect, photon, minimum energy, surface, frequency, eV, Joule.
