The threshold frequency fo of Cesium is about 9.39 × 10¹⁴ Hz
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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 )
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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!
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Complete Question:
What is the threshold frequency fo of Cesium?
Note that 1 eV (electron volt) = 1.60×10⁻¹⁹ J ?
[tex]\begin{tabular} {|c|c|c|}\underline{Light\ Energy\ (eV)}& \underline{Electron\ Emitted} & \underline{Electron KE\ (eV)}\\3.87 & no & -\\3.88 & no & -\\ \boxed{3.89} & yes & 0\\3.90 & yes & 0.01\\3.91 & yes & 0.02\end{tabular}[/tex]
Given:
the work function = Φ = 3.89 eV
Asked:
threshold frequency = fo = ?
Solution:
Firstly, we will convert the work function unit as follows:
Φ = 3.89 eV = 3.89 × 1.60 × 10⁻¹⁹ Joule = 6.224 × 10⁻¹⁹ Joule
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Next, we will calculate the threshold frequensy as follows:
[tex]\Phi = h \times fo[/tex]
[tex]6.224 \times 10^{-19} = 6.63 \times 10^{-34} \times fo[/tex]
[tex]fo = ( 6.224 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )[/tex]
[tex]{\boxed{fo \approx 9.39 \times 10^{14} ~ Hz}[/tex]
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Grade: High School
Subject: Physics
Chapter: Quantum Physics
