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 10^{8} m/s and Planck’s constant = 06.63 x 10^{-34} m^{2} 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 10^{8} /650 x 10^{-9}

= 4.6 x 10^{14} Hz

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

Energy E = 6.63 x 10^{-34} * 4.6 x 10^{14}

=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.

calculate the energy,in kJ/mole of a mole of photon in the infrared region of the electromagnetic spectrum that has a wavelength of 79.4 micro meter.

ReplyDelete