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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
$10^{-20}$
$10^{-17}$
$10^{-13}$
$10^{-10}$
Solution
$(c)$ 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
$\Rightarrow v ^{\prime} =h v +m g H$
$=h v +\frac{h v }{c^{2}} \cdot g H=h v\left(1+\frac{g H}{c^{2}}\right)$
So, $\frac{f^{\prime}-f}{f}=$ fractional change in
frequency $=\frac{g H}{c^{2}}=\frac{10 \times 1000}{\left(3 \times 10^{8}\right)^{2}}=\frac{10}{9} \times 10^{-13}$
