The photoelectric threshold wavelength for this copper surface is 264 nm
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:
λ = 254 nm = 2,54 × 10⁻⁷ m
V = 0.181 Volt
c = 3 × 10⁸ m/s
h = 6.63 × 10⁻³⁴ Js
q = 1.6 × 10⁻¹⁹ C
Unknown:
λ₀ = ?
Solution:
[tex]E = qV + \Phi[/tex]
[tex]h f = qV + h f_o[/tex]
[tex]h \frac{c}{\lambda} = qV + h \frac{c}{\lambda_o}[/tex]
[tex]6.63 \times 10^{-34} \times \frac{3 \times 10^8}{2.54 \times 10^{-7}} = 1.6 \times 10^{-19}(0.181) + 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{\lambda_o}[/tex]
[tex]7.83 \times 10^{-19} = 2.896 \times 10^{-20} + \frac{1.989 \times 10^{-25}}{\lambda_o}[/tex]
[tex]\frac{1.989 \times 10^{-25}}{\lambda_o} = 7.83 \times 10^{-19} - 2.896 \times 10^{-20}[/tex]
[tex]\lambda_o \approx 2.64 \times 10^{-7} ~ m[/tex]
[tex]\large {\boxed{\lambda_o \approx 264 ~ nm} }[/tex]
Grade: College
Subject: Physics
Chapter: Quantum Physics
Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light
