## Photon Energy Formula in Detail

Do you know what is a photon? Well!! Light emits a packet of energy, these chunks are photons and each photon carries its own energy, which is the photon energy.

The formula of energy of photon helps calculate the energy carried by each photon, which is given as:

E = hf….(1)

This energy of photon equation is also known as Planck-Einstein Relation.

Here,

E is the photon energy, i.e., the energy of a photon equation for a single photon

h = it is a constant, known as Planck’s constant.

f = the electromagnetic frequency, measured in Hertz or Hz

The above energy of a photon equation is valid for a single photon, if there are ‘n’ no of photons emitted by your tube light, then the no of photons formula is:

E = n* h* f

So, putting n = 1, we get the energy of one photon formula.

Here, energy ‘E’ is calculated in both J and eV, depending on the system of unit used.

On this page, we will understand the formula of photon energy, no of photons formula, photon wavelength formula, and the kinetic energy of photon formula.

### Photon Energy

We can express the photon energy by using any of the units of energy.

Among the units, we commonly use the electronvolt (eV) unit of the photon energy and the joule (it’s multiple, such as microjoule).

As 1 Joule = 6.24 × 10\[^{18}\] eV, however, the large units are often useful in denoting the photon energy with higher frequency and higher energy, such as gamma rays, as opposed to lower energy photons, like those in the radiofrequency region of the electromagnetic spectrum.

Now, let’s understand the photon energy formula in more detail:

### Formula of Photon Energy

The photon energy formula can be rewritten in the following way:

E = hf

Also, the energy photon formula frequency is c/λ. Putting the value of ‘f’ in the above equation:

E = hc/λ ….(2)

E is the photon energy in Joules

λ is the photon's wavelength in metres

c is the speed of light in a vacuum, whose value is 3 x 10\[^{8}\] metres per second

h is the Planck constant - Its value is 6.626 × 10\[^{-34}\] kgm\[^{2}\]s\[^{-1}\] or J.s

The photon energy at 1 Hz is equal to 6.626 × 10\[^{-34}\] J

Planck’s constant can be written in terms of eV, which is 4.14 × 10\[^{-15}\] eV· s.

### Energy of Photon Formula

From the energy of photon equation, we see that the energy of photon depends on the following parameters:

The energy of photon is directly related to the photon's electromagnetic frequency.

The energy of photon depends on wavelength in such a way that the energy of photon is inversely proportional to the wavelength.

The higher is the photon energy frequency, the higher its energy. In contrast, the longer is the photon’s wavelength, the lower its energy.

### Photon Wavelength Formula

We know that a photon is characterized by either a wavelength ‘λ’ or equivalently energy, ‘E’.

Also, there is an inverse relationship between E and λ, as stated in equation (2).

On multiplying the values of h and c as;

h * c = (6.626 × 10\[^{-34}\]J.s) * (2.998 × 10\[^{8}\]) = 1.99 × 10\[^{-25}\] J.m

The above inverse relationship helps us understand that the light of high-energy photons (such as "blue" light) has a short wavelength whereas the light of low-energy photons (such as "red" light) has a long wavelength.

While dealing with "particles" like photons, eV or the electronvolt is the most commonly used unit of energy than the joule (J). An electron volt is the energy needed to raise an electron through 1 volt, thus the photon energy in 1 eV = 1.602 × 10\[^{-19}\] J.

Therefore, the expression of hc in terms of eV will be:

hc = (1.99 × 10\[^{-25}\] J-m) × (1ev/1.602 × 10\[^{-19}\] J) = 1.24 × 10\[^{-6}\] eV-m

Further, writing the units for λ in terms of µm:

hc = (1.24 × 10\[^{-6}\] eV-m) × (10\[^{6}\] µm/ m) = 1.24 eV-µm

By expressing the energy of a photon equation in terms of eV and µm we arrive at a commonly used expression that relates the photon’s energy and wavelength, which we will understand under the further “energy of photon formula in eV” section.

### Energy of Photon Formula in eV

Energy is often measured in electronvolts.

The photon energy formula in electronvolts can be written by using the wavelength in micrometres, which is as follows:

E (eV) = \[\frac{1.2398}{\lambda (\mu m)}\]…..(3)

This equation is also known as the photon wavelength formula.

From equation (3), we observe that the exact value of 1 × 10\[^{6}\] (hc/q) is 1.24 but the approximation of 1.24 is sufficient for most purposes.

Also, this equation is significant for the wavelength in micrometres. From the above energy of photon formula in eV, we infer that photon energy at 1 μm wavelength, the wavelength of near-infrared radiation, is approximately 1.2398 eV or 1.24 eV.

### Kinetic Energy of Photon Formula

Since we know that the electrons are tightly bonded to the metal, so we need the energy to help them come out of the metal, i.e., to lead the photoemission process. So, the electrons that come out of the metal have some energy.

Therefore, the maximum kinetic energy of ejected electrons is shown below:

KE\[_{e}\] = hf - BE….(4)

Here,

E is the photon energy

BE = binding energy or the Work function of the electron, which is particular to the given material.

KE\[_{e}\] = kinetic energy (in Joules)

### Kinetic Energy vs Frequency Graph

The graph for the above equation is shown below:

[Image will be Uploaded Soon]

The above graph is between the kinetic energy of an ejected electron, KE\[_{e}\], and the frequency of electromagnetic radiation influencing a certain material.

There is a threshold or a limited frequency below which no electrons can eject because each photon interacting with its respective electron has insufficient energy to break it away or leave its lattice point.

Above all, the threshold energy, KE\[_{e}\] increases linearly with the frequency ‘f’, which remains consistent with the relation in equation (4).

Besides this, the slope of this line is ‘h.’ So, this data can be utilized to ascertain Planck’s constant experimentally.

Do you know that Einstein gave the first-ever successful explanation of the above data by proposing the notion of photons-quanta of EM radiation?

### Photon Applications

An FM radio station transmitting 100 MHz electromagnetic frequency releases photons with an energy of around 4.1357 × 10\[^{-7}\] eV. From the mass-energy equivalence), we understand that this amount of energy is approximately 8 × 10\[^{-13}\] times the electron's mass, i.e., 9.1 x 10\[^{-31}\] kg.

Gamma rays are considered of very high energy, They have photon energies of 100 GeV to over 1 PeV, i.e., from 10\[^{11}\] to 10\[^{15}\] eV, or 16 nanojoules to 160 microjoules. This relates to frequencies of 2.42 × 10\[^{25}\] to 2.42 × 10\[^{29}\] Hz.

### Conclusion

So, we have the following two formulas for the energy of photon calculator:

E (J) = hf = hc/λ

E (eV) = \[\frac{1.2398}{\lambda (\mu m)}\]

Q1: State One Application of Photon.

Ans: During photosynthesis, the chlorophyll molecules absorb red-light photons of 700 nm wavelength in photosystem I, equaling to the photon energy of approximately 2 eV or 3 x 10^{-19} J, which, in turn, equals 75 kBT.

Here, kBT is the thermal energy. At least 48 photons are required for the synthesis of a single glucose molecule from CO_{2} and H_{2}O with a chemical potential difference of 5 x 10^{-18} J and a maximum energy conversion efficiency of 35%.

Q2: How are Photons Used Today?

Ans: The photon is all familiar with elementary particles. Each photon travels with a speed of light, i.e., 3 x 108 m/s, which makes them greatly significant in quantum physics, along with electricity generation.

Like in the photoelectric effect, the energy of photons is transmitted through the given material, which helps in the photoemission. Thus, photoemission is the process of electrons coming out of the metal, these electrons are used in electricity generation.

Q3: How Do I Calculate Kinetic Energy?

Ans: In classical mechanics, we can calculate the kinetic energy (KE) as;

KE is equal to half of an object's mass (½ * m) multiplied by the square of the velocity.

For instance, if a body with a mass of 30 kg is moving at a velocity of 8 meters per second, then the kinetic energy is calculated as:

(1/2 * 30 kg) * 8 m/s^{2}, which is equivalent to 120 J