In the experiment of Pound and Rebka, a 14.4-keV gamma ray fell through a distance of 22.5 m near Earth’s surface. What are the change in frequency and the percentage change in frequency due to the gravitational redshift?

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In the experiment of Pound and Rebka, a 14.4-keV gamma ray fell through a distance of 22.5 m near Earth’s surface. What are the change in frequency and the percentage change in frequency due to the gravitational redshift?

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Step 1

Gravitational Redshift

Gravitational redshift is the change in frequency of a photon when it travels out of a gravitational well. According to the General Theory of Relativity, heavy objects bend the spacetime in such a way that it creates a well. The object is at the center of the well. When a photon travels out of this well, it tends to lose energy. This is known as gravitational redshift. 

Step 2

Given, the energy of the gamma ray E=14.4 keV=14.4×1.6×10-19 J

If f is the frequency of the photon, then

hf=Ef=Eh=14.4×1.6×10-196.626×10-34=3.477×1015 Hz

The gravitational redshift is given as

ff=ghc2

f is the change in frequency, g is the acceleration due to gravity, h is the height from which the photon has fallen and cis the velocity of light in free space. 

Thus

ff=ghc2f=ghfc2=9.8×22.5×3.477×10153×1082=8.518 Hz

Thus the percentage change in frequency

ff×100%=ghc2×100=9.8×22.53×1082×100=2.45×10-13 %

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