The energy of the photon emitted is about 1.60 × 10⁻²⁴ Joule
Further explanation
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 = 12.4 cm = 12.4 × 10⁻² 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}{12.4 \times 10^{-2}}[/tex]
[tex]E \approx 1.60 \times 10^{-24} ~ Joule[/tex]
Learn more
- Photoelectric Effect : https://brainly.com/question/1408276
- Statements about the Photoelectric Effect : https://brainly.com/question/9260704
- Rutherford model and Photoelecric Effect : https://brainly.com/question/1458544
Answer details
Grade: College
Subject: Physics
Chapter: Quantum Physics
Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light
