How to calculate the energy of a photon?

You would remember from a previous post that a photon is a packet of energy. Hence all electromagnetic radiations are emitted in term of photons.

 

We also saw that the energy of a photon is given by the equation

E = hf

where h is the Planck’s constant and f is the frequency of the electromagnetic radiation’s photon.

Fig 1

From the equation we can thus deduce that a photon’s energy is directly proportional to its frequency.

Hence the higher the frequency of a photon the greater the energy it has.

From fig 1 we can deduce that since radio wave has the least frequency then the photons of radio waves has the least energy. And as shown in fig 1 as we move from left to right across the electromagnetic spectrum the frequency increases. This means that the energy of the photon also increases. Photons of gamma rays would thus be the most energetic.

Example

Calculate the energy of a photon of red light of wavelength 650 nm if the speed of light is 3.0 x 10⁸ m/s and Planck’s constant = 06.63 x 10^-34 m² kg s^-1.

Now if you remember the frequency f, the wavelength λ and the speed  c of all electromagnetic radiations are related by the equation

c = fλ

hence making f the subject of formula

f =c/λ

Calculating f

f = 3.0 x 10⁸ /650 x 10^-9

= 4.6 x 10¹⁴ Hz

Calculating the energy of the photon using the equation E = hf

Energy E = 6.63 x 10^-34 * 4.6 x 10¹⁴

                     =3.0 x10^-19 J

You would remember in a previous post that we discussed about the use of the electron volt eV as unit of energy in quantum mechanics.

We are now going to see how to determine the energy of the photon in eV

E = 3.0 x10^-19 J

= 3.0 x10^-19 /1.6 x10^-19

= 1.9 eV

As we have seen in the previous post working with eV is much better.

What is the electron volt (eV)?

The electron volt is a unit of energy like the joule (J).

However in quantum mechanics if the joule is used, then very small numbers would be obtained. These small numbers are very clumsy to use as a result a better unit to use is the electron volt.  

Fig 1

As you can see in fig 1, we have two metal plates that are connected to a power supply of e.m.f. 1 V. As a result a potential difference of 1 V will be set up be set up between the plates. An electric field will be set up between the plates as a result of the potential difference.

Now if a stationary electron is released from the negatively charged plate it will experience an acceleration due to the electric field and as a result will move towards the positively charged plate with increasing velocity.

When it would reached the positively charged plate it would have a certain velocity.

The Kinetic energy gained by the electron can be calculated by the following equation:

Kinetic energy gained E_k = Charge on particle * Potential difference between plates.

E_k = Q * V

Hence it can be deduce that if the charge is an electron then the kinetic energy that is gained by the electron is

Kinetic energy gained E_k = Charge on electron (elementary charge) * 1 V

   = 1 eV

Hence the eV is the energy that is gained by an electron or any particle of charge -e or +e when accelerated by a potential difference of 1 V 

If you want to know how many joule there is in 1 eV then it is as shown below

Kinetic energy gained E_k = Charge on electron (elementary charge) * 1 V

= 1.6 x 10^-19* 1

= 1.6 x10^-19 J

We can thus say that 1 eV is equivalent to 1.6 x 10 ^–19 J

As we have said above in quantum mechanics it is better to use the the eV because if the joule is used we would be dealing with small numbers which would make the works and calculations difficult.