The frequency of the photon is 8.34 × 10¹⁶ 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:
E = 5.53 × 10⁻¹⁷ m
h = 6.63 × 10⁻³⁴ Js
Unknown:
f = ?
Solution:
[tex]E = h \times f[/tex]
[tex]5.53 \times 10^{-17} = 6.63 \times 10^{-34} \times f[/tex]
[tex]f = 5.53 \times 10^{-17} \div (6.63 \times 10^{-34})[/tex]
[tex]f \approx 8.34 \times 10^{16} ~ Hz[/tex]
Grade: College
Subject: Physics
Chapter: Quantum Physics
Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light
