The energy of exactly one photon of this microwave radiation is [tex]\boxed{1.584\times10^{-24}\text{ J}}[/tex].
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 [tex]3.0\times{10^8}{\text{ m/s}}[/tex] 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 [tex]12.5\text{ m}[/tex].
Concept:
The expression for the energy of photon is:
[tex]\boxed{E=\dfrac{{h\cdot c}}{\lambda}}[/tex]
Here, [tex]E[/tex] is the energy of photon, [tex]h[/tex] is the Plank constant, [tex]c[/tex] is the speed of the light and [tex]\lambda[/tex] is the wavelength of the photon.
The value of the Plank constant is [tex]6.6\times{10^{-34}}{\text{ J}\cdot\text{s}}[/tex].
The value of the speed of the light is [tex]3.0\times{10^8}\text{ m/s}[/tex].
Substitute [tex]6.6\times{10^{-34}}{\text{ J}\cdot\text{s}}[/tex] for [tex]h[/tex], [tex]3.0\times{10^8}\text{ m/s}[/tex] for [tex]c[/tex] and [tex]12.5\text{ cm}[/tex] for [tex]\lambda[/tex] in the above equation.
[tex]\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}[/tex]
Therefore, the energy of exactly one photon of this microwave radiation is [tex]\boxed{1.584 \times {10^{ - 24}}{\text{ J}}}[/tex]
Learn More:
1. The threshold frequency of the cesium https://brainly.com/question/6953278
2. The direction of propagation of a sound wave https://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.
