Photon Theory Of Light MCQ - Practice Questions with Answers

Photon theory of light -

According to Eienstein's quantum theory light propagates in the bundles (packets or quanta) of energy, each bundle being called a photon and possessing energy. The energy of one quantum is given by, hν , where h is the Planck's constant and ν is the frequency.

$$E=h\nu =\frac{hc}{\lambda }$$

where $\mathrm{c}=$ Speed of light, $\mathrm{h}=$ Plank's constant $=6.6\times {10}^{-34}\mathrm{J}-\sec$
$\nu =$ Frequency in $\mathrm{Hz},\lambda =$ Wavelength of light.

$$E(eV)=\frac{1.24}{\lambda (\mu m)}$$

(In the form of eV)
Photon is a packet of energy (or) particles of light and travels with speed light in a straight line. Energy of photon depends on frequency and it does not change with change in medium. One more thing is that the Photons are electrically neutral and not affected by electronic and

                           (In the form of eV)

Photon is a packet of energy (or) particles of light and travels with speed light in a straight line. Energy of photon depends on frequency and it does not change with change in medium. One more thing is that the Photons are electrically neutral and not affected by electronic and magnetic field. 

After energy now let us discuss the mass of the photon.

Mass of photon : 

You will study in the theory of relativity that the rest mass of any body is given by - 

$$m_v=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$\text{ Where: }\begin{array}{rl}{m}_v& =\text{ Relativistic mass }(\mathrm{kg})\\ m_o& =\text{ Rest mass }(\mathrm{kg})\\ v& =\text{ velocity }\left({\mathrm{ms}}^{-1}\right)\\ \mathrm{c}& =\text{ speed of light }=3\times {10}^8{\mathrm{ms}}^{-1}\end{array}$$

As the velocity of photon is same as speed of light, so from the above equation we can write that - $m_o=0$. But it's effective mass is given as -

$$E=m{c}^2=h\nu \Rightarrow m=\frac{E}{c^2}=\frac{h\nu }{c^2}=\frac{h}{c\lambda }$$

It is also called as kinetic mass of the photon.
Momentum of the photon -
As the momentum of any body is $=m\cdot v$
Here the velocity $=c$, i.e., speed of light. So, we can write that -

$$p=m\times c=\frac{E}{c}=\frac{h\nu }{c}=\frac{h}{\lambda }$$
 

Note -

1. In a photon particle collision, total energy and total momentum will be conserved but the number of photons may be changed. 

2. All photons of light of a particular freq. (or) wavelengths have the same energy and momentum whatever may be the intensity.