Answer:
6.383 x 10^14 Hz
Explanation:
we know that all the electromagnetic radiations moves with the velocity of light.
the velocity of light is 3 x 10^8 m/s.
The relation between the wave velocity, frequency and the wavelength is given by
v = f λ
Where, λ is wavelength and f be the frequency
f = v / λ
f = (3 x 10^8) / (470 x 10^-9)
f = 6.383 x 10^14 Hz
The frequency of Blue Light is 6.40 × 10¹⁴ Hz
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 = 470 nm = 470 × 10⁻⁹ m
Unknown:
f = ?
Solution:
[tex]f = \frac{c}{\lambda}[/tex]
[tex]f = \frac{3 \times 10^8}{470 \times 10&{-9}}[/tex]
[tex]\large {\boxed {f \approx 6.40 \times 10^{14} ~ Hz} }[/tex]
Grade: High School
Subject: Physics
Chapter: Quantum Physics
Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt
