
How does photon energy change with increasing frequency?
Answer
524.7k+ views
Hint: The relation for the energy of the photon and frequency of the photon can be explained from Plank's Quantum theory that states the formula $E=hv$, where E is the energy of the photon, h is the Planck's constant, and $v$ is the frequency.
Complete answer:
The relation between the energy of the photon and the frequency of the photon can be explained from Plank's Quantum theory. He stated this theory for the explanation of 'Black body radiations' and 'Photoelectric effect', in which he stated that the radiant energy is emitted or absorbed discontinuously not continuously in discrete packets called 'quantum' or 'photon'. He has also given a formula of the relation between the energy and frequency:
$E=hv$
Where E is the energy of the photon, h is the Planck's constant, and $v$ is the frequency.
Since h is a constant, when it is removed, then the energy of the photon is directly proportional to the frequency of the photon.
\[E\propto v\]
So, as the frequency of the photon will increase, the energy of the photon will also increase.
Additional information:
If we have to relate with the energy of the photon and wavelength of the photon, then we have to use the formula:
$E=\dfrac{hc}{\lambda }$
Where E is the energy, h is the planck's constant, c is the velocity of light and $\lambda $ is the wavelength. Since h and c are constants, then we can say that the energy of the photon is inversely proportional to the wavelength of the photon. So, as the energy increases, the wavelength decreases.
Note:
The total amount of energy emitted or absorbed by a body, can never be a fractional value or negative, it can only be an integer value, so the formula is written as:
$E=nhv$
Where n is an integer.
Complete answer:
The relation between the energy of the photon and the frequency of the photon can be explained from Plank's Quantum theory. He stated this theory for the explanation of 'Black body radiations' and 'Photoelectric effect', in which he stated that the radiant energy is emitted or absorbed discontinuously not continuously in discrete packets called 'quantum' or 'photon'. He has also given a formula of the relation between the energy and frequency:
$E=hv$
Where E is the energy of the photon, h is the Planck's constant, and $v$ is the frequency.
Since h is a constant, when it is removed, then the energy of the photon is directly proportional to the frequency of the photon.
\[E\propto v\]
So, as the frequency of the photon will increase, the energy of the photon will also increase.
Additional information:
If we have to relate with the energy of the photon and wavelength of the photon, then we have to use the formula:
$E=\dfrac{hc}{\lambda }$
Where E is the energy, h is the planck's constant, c is the velocity of light and $\lambda $ is the wavelength. Since h and c are constants, then we can say that the energy of the photon is inversely proportional to the wavelength of the photon. So, as the energy increases, the wavelength decreases.
Note:
The total amount of energy emitted or absorbed by a body, can never be a fractional value or negative, it can only be an integer value, so the formula is written as:
$E=nhv$
Where n is an integer.
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