The energy of individual photons of 450 nm light is 4.42 × 10⁻¹⁹ 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 = 450 nm = 450 × 10⁻⁹ m
Unknown:
E = ?
Solution:
[tex]E = h f[/tex]
[tex]E = h \frac{c}{\lambda}[/tex]
[tex]E = 6.63 \times 10^{-34} \frac{3 \times 10^8}{450 \times 10^{-9}}[/tex]
[tex]\large {\boxed {E = 4.42 \times 10^{-19} ~ 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
- Photoelectric Threshold Wavelength : https://brainly.com/question/10015690
Answer details
Grade: High School
Subject: Physics
Chapter: Quantum Physics
Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt
