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Question

(C) photon of mass $m$ and frequency $
u$ falls through a height $H$ in the earth's gravitational field.When the photon falls through the earth's gra

Vedclass · Accepted Answer

$10^{-13}$

Vedclass · Answer

(C) photon of mass $m$ and frequency $
u$ falls through a height $H$ in the earth's gravitational field.When the photon falls through the earth's gravitational field,it gains extra energy. Therefore,Final photon energy = Initial photon energy + Increase in energy$h 
u' = h 
u + m g H$Substituting $m = \frac{h 
u}{c^{2}}$,we get:$h 
u' = h 
u + \left( \frac{h 
u}{c^{2}} ight) g H$$
u' = 
u \left( 1 + \frac{g H}{c^{2}} ight)$So,the fractional change in frequency is:$\frac{
u' - 
u}{
u} = \frac{g H}{c^{2}}$Given $g = 10 \, m/s^{2}$,$H = 1000 \, m$,and $c = 3 	imes 10^{8} \, m/s$:$\frac{\Delta 
u}{
u} = \frac{10 	imes 1000}{(3 	imes 10^{8})^{2}} = \frac{10^{4}}{9 	imes 10^{16}} \approx 1.11 	imes 10^{-13}$This value is close to $10^{-13}$.

$A$ photon falls through a height of $1 \, km$ in the earth's gravitational field. To calculate the change in its frequency,take its mass to be $h \nu / c^{2}$. The fractional change in frequency $\nu$ is close to

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