A hydrogen atom, when vigorously perturbed, can emit light with a frequency of 6.16 × 1014 Hz. When the same light from hydrogen atoms in a distant galaxy is observed on earth, the frequency is 5.21 × 1014 Hz. Calculate the speed at which the galaxy is receding from the earth (in units of the speed of light, c).
Since v is much less than c, you can use the approximate formula
(delta f)/fo = 0.95*10^14/6.16*10^14
= 0.154 = v/c
v = 0.154 c = 4.6*10^7 m/s
There is an exact formula for the relativistic Doppler shift of light, but I don't believe they expect you to know it. You can find it at
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop2.html
To calculate the speed at which the galaxy is receding from Earth, we can use the formula for the redshift, which relates the observed frequency to the emitted frequency:
z = (observed frequency - emitted frequency) / emitted frequency
where z is the redshift. In this case, the observed frequency is 5.21 × 10^14 Hz and the emitted frequency is 6.16 × 10^14 Hz.
Plugging in the values, we have:
z = (5.21 × 10^14 Hz - 6.16 × 10^14 Hz) / 6.16 × 10^14 Hz
z ≈ -0.171
The redshift, z, is related to the velocity of the galaxy using Hubble's law:
z = v / c
where v is the velocity of the galaxy and c is the speed of light.
Rearranging the equation, we can solve for v:
v = z * c
Plugging in the value for z, we get:
v = -0.171 * c
Therefore, the speed at which the galaxy is receding from the Earth is approximately -0.171 times the speed of light, c.