Microwave ovens emit microwave energy with a wavelength of 12.5 cm. What is the energy of exactly one photon of this microwave r

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Microwave ovens emit microwave energy with a wavelength of 12.5 cm. What is the energy of exactly one photon of this microwave r

adiation?
I'm on sapling and I did it couple times, but I kept getting it wrong!!! I followed all the notes I have from lecture, can someone please help

thank you

Physics

2 answers:

enyata [817] · Answer 1 · 2021-11-23T22:09:00+03:00

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, $E$ is the energy of photon, $h$ is the Plank constant, $c$ 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 $h$ , $3.0\times{10^8}\text{ m/s}$ for $c$ 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 brainly.com/question/6953278

2. The direction of propagation of a sound wave brainly.com/question/3619541

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.

Andrej [43] · Answer 2 · 2021-11-23T21:05:06+03:00

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

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

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

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

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.

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