The maximum kinetic energy is about 2.59 eV
[tex]\texttt{ }[/tex]
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 )
[tex]\texttt{ }[/tex]
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 !
[tex]\texttt{ }[/tex]
Given:
wavelength of first light = λ₁ = 400 nm = 4 × 10⁻⁷ m
the first maximum kinetic energy = Ek₁ = 1.10 eV = 1.76 × 10⁻¹⁹ J
wavelength of second light = λ₂ = 270 nm = 2.7 × 10⁻⁷ m
Asked:
the second maximum kinetic energy = Ek₂ = ?
Solution:
[tex]E_1 = Ek_1 + \Phi[/tex]
[tex]h \frac{c}{\lambda_1} = Ek_1 + \Phi[/tex]
[tex]h \frac{c}{\lambda_2} = Ek_2 + \Phi\\\texttt{\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ }} ( - )[/tex]
[tex]hc ( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} }) = Ek_1 - Ek_2[/tex]
[tex]6.63 \times 10^{-34} \times 3 \times 10^8 \times ( \frac{1}{4 \times 10^{-7}} - \frac{1}{2.7 \times 10^{-7}} ) = 1.76 \times 10^{-19} - Ek_2[/tex]
[tex]- 2.39 \times 10^{-19} = 1.76 \times 10^{-19} - Ek_2[/tex]
[tex]Ek_2 = 1.76 \times 10^{-19} + 2.39 \times 10^{-19}[/tex]
[tex]Ek_2 = 4.15 \times 10^{-19} \texttt{ J}[/tex]
[tex]Ek_2 = 2.59 \texttt{ eV}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Quantum Physics
