Respuesta :
Explanation:
Given that,
The wavelength of high‑frequency radio wave, [tex]\lambda=0.25\ km=250\ m[/tex]
We need to find the energy of exactly one photon of this radio wave radiation. It is given by :
[tex]E=\dfrac{nhc}{\lambda}[/tex]
Here, n = 1
[tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{250}\\\\E=7.95\times 10^{-28}\ J[/tex]
So, the energy of exactly one photon of this radio wave radiation is [tex]7.95\times 10^{-28}\ J[/tex].
The energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].
Given that,
- The equipment used for aviation communications emits high‑frequency radio wave energy with a wavelength of 0.250 km i.e. = 250 m.
Based on the above information, the calculation is as follows:
We know that
[tex]E = nhc \div \lambda\\\\= (6.63 \times 10^{-34} \times 3 \times 10^8) \div 250[/tex]
= [tex]7.95 \times 10^{-28}J[/tex]
Therefore we can conclude that the energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].
Learn more: brainly.com/question/23334479
