Answer:
[tex]E = 3.36 \times 10^{-19} J[/tex]
Explanation:
as we know that threshold frequency is the minimum frequency of light at which electrons are ejected out of the metal surface
So here we know that
[tex]E = h\nu[/tex]
so we have
[tex]h = 6.6 \times 10^{-34} Js[/tex]
now we have
[tex]\nu = 5.09 \times 10^{14} Hz[/tex]
so we have
[tex]E = (6.6 \times 10^{-34})(5.09 \times 10^{14})[/tex]
[tex]E = 3.36 \times 10^{-19} J[/tex]
The minimum energy needed is about 3.37 × 10⁻¹⁹ Joule
[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:
threshold frequency = fo = 5.09 × 10¹⁴ Hz
Asked:
minimum energy = Φ = ?
Solution:
[tex]\Phi = h f_o[/tex]
[tex]\Phi = 6.63 \times 10^{-34} \times 5.09 \times 10^{14}[/tex]
[tex]\Phi = 3.37 \times 10^{-19} \texttt{ Joule}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Quantum Physics
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Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light
