Question

According to the quantum theory, a photon of electromagnetic radiation of frequency $v$ has energy $E=hv$ where $h$ is known as planck's constant. According to the theory of relativity, a particle of mass m has equivalent energy $E$ = $m{c}^2$, where $c$ is speed of light. Thus a photon can be treated as a particle having effective mass $m=\frac{hv}{c^2}$. If a flash of light is sent horizonatally in earth's gravitational field, then photons while traveling a horizontal distance $d$ would fall through a distance given by

• A. $\frac{g{d}^2}{2c^2}$
• B. $\frac{h}{mc}$
• C. $\frac{mc{d}^2}{h}$
• D. Zero

Answer 1

Answer: (A)

$\frac{g{d}^2}{2c^2}$

$E={mc}^2=h\upsilon$

$\therefore m=\frac{h\upsilon }{c^2}$ (Given)

Here, the P.E of the photons due to gravitaional force is given by:

$P.E=\frac{GMm}{2D}=\frac{g{d}^{2}m}{2D}$$(\therefore GM=g{d}^2)$

M= mass of earth

m= mass of particle

$P.E={mc}^2=\frac{g{d}^{2}m}{2D}$

$\therefore \frac{g{d}^2}{2c^2}=D$

Where $D=$ distance at which particle will fall.