Question

Microwave ovens emit microwave energy with a wavelength of 12.6 cm. what is the energy of exactly one photon of this microwave radiation?

Answer 1

We need the frequency of the photon, it is v = c/ λ

Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon 

which is E = hv where h is Planck’s constant

Combining the two equations will give us:

E = h x c/λ

Inserting the values, we will have:

E = 6.626 x 10^-34 x 3 x 10^8 / 0.126

E = 1.578 x 10^ -24 J

Answer 2

The energy of exactly one photon of this microwave radiation is about 1.58 × 10⁻²⁴ J

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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 :

$\large {\boxed {E = h \times f}}$

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.

$\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}$

$\large {\boxed {E = qV + \Phi}}$

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:

c = 3 × 10⁸ m/s

h = 6.63 × 10⁻³⁴ Js

λ = 12.6 cm = 0.126 m

Unknown:

E = ?

Solution:

$E = h \times f$

$E = h \times \frac{c}{\lambda}$

$E = (6.63 \times 10^{-34}) \times \frac{3 \times 10^8}{0.126}$

$E = 1.58 \times 10^{-24} \texttt{ J}$

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Learn more

• Photoelectric Effect : brainly.com/question/1408276
• Statements about the Photoelectric Effect : brainly.com/question/9260704
• Rutherford model and Photoelecric Effect : brainly.com/question/1458544

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Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics