Answer:
[tex]E=4.847x10^{-19}J[/tex]
Explanation:
Using the equation for energy:
[tex]E=\frac{hc}{\lambda}[/tex]
where h is plack's constant:
[tex]h=6.626x10^{-34}J*s[/tex]
and c is the velocity of light:
[tex]c=3x10^8m/s[/tex]
and [tex]\lambda[/tex] is the wavelength of the light:
[tex]\lambda=410.1 nm=410.1 x10^{-9}m[/tex]
Thus, the energy E:
[tex]E=\frac{(6.626x10^{-34}J*s)(3x10^8m/s)}{410.1 x10^{-9}m}[/tex]
[tex]E=4.847x10^{-19}J[/tex]
The energy of the violet light emitted is about 4.85 × 10⁻¹⁹ Joule
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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 )
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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 !
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Given:
λ = 410.1 nm = 4.101 × 10⁻⁷ m
h = 6.63 × 10⁻³⁴ Js
c = 3 × 10⁸ m/s
Unknown:
E = ?
Solution:
[tex]E = h \times \frac{c}{\lambda}[/tex]
[tex]E = 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{4.101 \times 10^{-7}}[/tex]
[tex]E \approx 4.85 \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
