Question

If the wavelength of electromagnetic radiation is doubled, what happens to the energy of the photon?

Answer 1

Hint: When there is no net flow of matter or energy between the body and its surroundings, Planck's law defines the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T. Physicists couldn't explain why the observed spectrum of black-body radiation, which had been properly measured at that time, varied substantially at higher frequencies from that predicted by current theories at the end of the 19th century.

Complete step by step solution:
Max Planck proposed the notion that energy is transmitted in the form of quanta, which he designated as h. The variable h represents the frequency in\[{{s}^{-1}}\] and has a constant value of \[6.63\text{ }x\text{ }{{10}^{-34}}\text{ }J.s\]based on the International System of Units. When the frequency of photons is known, we can use Planck's law to determine their energy. If you know the wavelength, you can determine the energy by first calculating the frequency with the wave equation and then using Planck's equation to obtain the energy. To put it another way, Plank's constant defines the relationship between the energy per quantum (photon) and the frequency of electromagnetic radiation.
$E=hv=h\dfrac{c}{\lambda }$
Hence
$E\alpha \dfrac{1}{\lambda }$
As a result, doubling the wavelength reduces the photon's energy by half.

Note: Every physical body produces electromagnetic radiation spontaneously and continuously, and a body's spectral radiance, B, defines the spectral emissive power per unit area, per unit solid angle for certain radiation frequencies. Planck's radiation law, as seen below, demonstrates that when a body's temperature rises, its total radiated energy rises as well, and the peak of the emitted spectrum moves to shorter wavelengths.