Answer:
[tex]9.37\cdot 10^{-19} J[/tex]
Explanation:
The material is irradiated with light of wavelength
[tex]\lambda=120 nm=1.2\cdot 10^{-7}m[/tex]
So the energy of the incoming photons is
[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.2\cdot 10^{-7} m}=1.66\cdot 10^{-18} J[/tex]
The threshold frequency for the emission of photoelectrons in the material is
[tex]f=1.09\cdot 10^{15}s^{-1}[/tex]
which corresponds to an energy of
[tex]E_{th}=hf=(6.63\cdot 10^{-34} Js)(1.09\cdot 10^{15} Hz)=7.23\cdot 10^{-19} J[/tex]
Therefore, the maximum possible kinetic energy of the emitted electrons will be equal to the difference between the energy of the incoming radiation and the threshold energy:
[tex]E=1.66\cdot 10^{-18} J-7.23\cdot 10^{-19}J=9.37\cdot 10^{-19} J[/tex]
The maximum kinetic energy is about 9.35 × 10⁻¹⁹ J
[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:
minimum frequency = fo = 1.09 × 10¹⁵ Hz
wavelength of light = λ = 120 nm
Asked:
maximum possible kinetic energy = Ek = ?
Solution:
[tex]E = Ek + hf_o[/tex]
[tex]h \frac{c}{\lambda} = Ek + hf_o[/tex]
[tex]Ek = h \frac{c}{\lambda} - hf_o[/tex]
[tex]Ek = (6.63 \times 10^{-34} \times \frac{3 \times 10^8}{120 \times 10^{-9}}) - (6.63 \times 10^{-34} \times 1.09 \times 10^{15})[/tex]
[tex]Ek \approx 9.35 \times 10^{-19} \texttt{ J}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Quantum Physics
