Let us discuss some questions involving photons and electrons. The following question high lights one basic difference between photons and electrons:

**The kinetic energies of an electron and a photon are in the ratio 9:4. Their momenta are in the ratio**

(a) 9:4 (b) 3:2 (c) 1:3 (d) 4:3 (e) 3:4

The kinetic energy of an electron (or any such particles of matter) is directly proportional to the square of its momentum (since E = p^{2} /2m with usual notations) where as the kinetic energy of a photon is directly proportional to its momentum (since E = mc^{2} = mc.c = pc). The ratio of the momenta of the electron and the photon is therefore √9: 4 = 3:4 [Option (e)].

You will encounter many situations in which you will be required to calculate the velocity of a particle of charge ‘q’ and mass ‘m’ accelerated by a voltage ‘V’. The equation you have to use is ½ mv^{2} = qV from which the velocity, v = √(2qV/m).

A special case is that of the electron, which you will encounter quite often. When you substitute the mass and the charge of the electron, the velocity is given (approximately) by v = 6×10^{5}√V. Remember this relation. You can save the valuable time you will have to spend on certain calculations. For instance, consider the following M.C.Q.:

**An electron at rest is accelerated by 2500 volts. Its final velocity is approximately (in m/s)**

(a) 3×10^{6} (b) 3×10^{7 }(c) 2.5×10^{8 }(d) 2.5×10^{6 }(e) ) 1.5×10^{8}

If you use the relation, v = 6×10^{5}√V , you will get the correct option (b) in no time.

[You should note that we have not considered the relativistic increase in the mass of the electron in the above approximate equation. If the accelerating voltage is higher, the relativistic increase in the mass is significant and the above equation cannot be used. Even in the above problem, the velocity of the electron is 10% of the velocity of light and the relativistic increase in the mass begins to be exhibited].

Now consider the following question:

**A laser source has output power of 300 mW at a wave length of 6630Ǻ. The number of photons emitted from this laser source every minute is**

(a) 6×10^{19 }(b) 6×10^{18 }(c) 6×10^{17 }(d) 6×10^{16} (e) 6×10^{14}

If the number of photons emitted per second is ‘n’, we have nhν = nhc/λ = 300×10-3 so that n = 0.3 λ/hc where ‘h’ is Planck’s constant, ‘c’ is the speed of light and ‘λ’ is the wave length. The number of photons emitted per minute is 60n = (60×0.3×6630×10^{-10})/(6.63×10^{-34}×3×10^{8}) = 6×10^{19} [Option (a)].

**Energy and Wave Length of a Photon**

There are many situations in which you will have to find the energy of a photon in electron volt if its wave length in Angstrom Units (A.U.) is given and vice versa. For instance, suppose you are asked to find out the minimum wave length of X-rays that can be obtained from an X-ray tube operating with an anode voltage of 20kV. Since the X-ray photon of minimum wave length will have the entire energy of 20000 electron volt (of the electron striking the target in the X-ray tube), you can write,20000 e = hc/λ where ‘e’ is the electronic charge, ‘h’ is Planck’s constant, ‘c’ is the speed of light in free space and λ is the wave length in metre.Instead of using the above equation for calculating λ, you may remember that in the case of a photon,λE = 12400 where λ is in Angstrom and the energy E is in electron volt.The answer to our present problem therefore is,

λ = 12400/20000 = 0.62 A.U.

Do you know *why germanium and silicon are unsuitable for making light-emitting diodes?*

The band gap for producing visible light should be at least 12400/7000 electron volt (since the maximum wave length of visible light is roughly 7000 Angstrom). This works out to be nearly 1.8eV, which is greater than the band gap of Ge and Si.