A radiation of energy $$'E'$$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ($$c$$ = velocity of light)
A.
$$\frac{E}{c}$$
B.
$$\frac{{2E}}{c}$$
C.
$$\frac{{2E}}{{{c^2}}}$$
D.
$$\frac{E}{{{c^2}}}$$
Answer :
$$\frac{{2E}}{c}$$
Solution :
The radiation energy is given by $$E = \frac{{hc}}{\lambda }$$
Initial momentum of the radiation is
$${P_i} = \frac{h}{\lambda } = \frac{E}{c}$$
The reflected momentum is
$${P_r} = - \frac{h}{\lambda } = - \frac{E}{c}$$
So, the change in momentum of light is
$$\Delta {P_{{\text{light}}}} = {P_r} - {P_i} = - \frac{{2E}}{c}$$
Thus, the momentum transferred to the surface is
$$\Delta {P_{{\text{light}}}} = \frac{{2E}}{c}$$
Releted MCQ Question on Modern Physics >> Dual Nature of Matter and Radiation
Releted Question 1
A particle of mass $$M$$ at rest decays into two particles of
masses $${m_1}$$ and $${m_2},$$ having non-zero velocities. The ratio of the de Broglie wavelengths of the particles, $$\frac{{{\lambda _1}}}{{{\lambda _2}}},$$ is
A proton has kinetic energy $$E = 100\,keV$$ which is equal to that of a photon. The wavelength of photon is $${\lambda _2}$$ and that of proton is $${\lambda _1}.$$ The ration of $$\frac{{{\lambda _2}}}{{{\lambda _1}}}$$ is proportional to