A photon falls through a height of 1 ,km thro | vedclass

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Question

$(c)$ A photon of mass $m$ and frequency $v$ falls through a height $H$ through the carth's gravitational ficld as shown in figure below.

When

Vedclass · Accepted Answer

$10^{-13}$

Vedclass · Answer

$(c)$ A photon of mass $m$ and frequency $v$ falls through a height $H$ through the carth's gravitational ficld as shown in figure below.

When the photon falls through the earth's gravitational field, then it is gains extra energy. Therefore,

Final photon energy

$=$ Initial photon energy $+$ Increase in energy

$\Rightarrow v ^{\prime} =h v +m g H$

$=h v +\frac{h v }{c^{2}} \cdot g H=h v\left(1+\frac{g H}{c^{2}}ight)$

So, $\frac{f^{\prime}-f}{f}=$ fractional change in

frequency $=\frac{g H}{c^{2}}=\frac{10 	imes 1000}{\left(3 	imes 10^{8}ight)^{2}}=\frac{10}{9} 	imes 10^{-13}$

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

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