Physics, 13.01.2020 02:31 coleian24

Microwave ovens emit microwave energy with a wavelength of 12.5 cm. what is the energy of exactly one photon of this microwave radiation?

i'm on sapling and i did it couple times, but i kept getting it ! i followed all the notes i have from lecture, can someone

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Answer from: ldibut2003

The energy of exactly one photon of this microwave radiation is 1.59 × 10⁻²⁴Joule

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 )

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!

Given:

λ = 12.5 cm = 12.5 × 10⁻² m

Unknown:

E = ?

Solution:

E = h f

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

$E = 6.63 \times 10^{-34} \frac{3 \times 10^8}{12.5 \times 10^{-2}}$

$\large {\boxed {E = 1.59 \times 10^{-24} ~ Joule } }$

Learn morePhotoelectric Effect : Statements about the Photoelectric Effect : Rutherford model and Photoelecric Effect : Photoelectric Threshold Wavelength : Answer details

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

Microwave ovens emit microwave energy with a wavelength of 12.5 cm. what is the energy of exactly on

Answer from: sylvia47

The energy of exactly one photon of this microwave radiation is $\boxed{1.584\times10^{-24}\text{ J}}$ .

Further Explanation:

Planck’s equation used to solve this question. In the Planck’s equation, the energy of a photon is directly related to the frequency of a photon and inversely related to the wavelength of a photon.

By using Planck’s equation the energy of the one photon can be determined.

The term “Electromagnetic wave radiation energy” was first introduced by the scientist Max Planck.

The light can travel very fast as in the no other thing can travel as faster as light and it’s measures value is approximate $3.0\times{10^8}{\text{ m/s}}$ in vacuum.

The wavelength is the defined as the distance between two consecutive positive peak pointornegative peak point of the wave.

Given:

The wavelength of the energy is $12.5\text{ m}$ .

Concept:

The expression for the energy of photon is:

$\boxed{E=\dfrac{{h\cdot c}}{\lambda}}$

Here, is the energy of photon, is the Plank constant, is the speed of the light and $\lambda$ is the wavelength of the photon.

The value of the Plank constant is $6.6\times{10^{-34}}{\text{ J}\cdot\text{s}}$ .

The value of the speed of the light is $3.0\times{10^8}\text{ m/s}$ .

Substitute $6.6\times{10^{-34}}{\text{ J}\cdot\text{s}}$ for , $3.0\times{10^8}\text{ m/s}$ for and $12.5\text{ cm}$ for $\lambda$ in the above equation.

$\begin{aligned}E&=\dfrac{{\left( {6.6 \times {{10}^{ - 34}}{\text{ Js}}} \right) \cdot \left( {3.0 \times {{10}^8}{\text{ m/s}}} \right)}}{{12.5{\text{ cm}}}}\\&=\dfrac{{\left( {6.6 \times {{10}^{ - 34}}{\text{ Js}}} \right) \cdot \left( {3.0 \times {{10}^8}{\text{ m/s}}} \right)}}{{12.5 \times {{10}^{ - 3}}{\text{ m}}}}\\&=1.584 \times {10^{ - 24}}{\text{ J}}\\\end{aligned}$

Therefore, the energy of exactly one photon of this microwave radiation is $\boxed{1.584 \times {10^{ - 24}}{\text{ J}}}$

Learn More:

1. The threshold frequency of the cesium

2. The direction of propagation of a sound wave

Answer Details:

Grade: High School

Subject: Physics

Chapter: Photoelectric effect

Keywords:

The energy, photon, microwave, radiation, emits, wavelength ,12.5 cm or 0.125 m, 1.59x10^-23 J, speed of light, emission, Plank's constant.

Answer from: Quest

The answer is 12, for the points

Answer from: Quest

Velocity is the speed of an object in a certain direction so you'd need to know speed and direction

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