ANSWER
6.4 x 10¹⁶ Hz
EXPLANATION
Given:
• The speed of the source, vs = 236 km/s
,• The speed of light, c = 300,000 km/s
,• The observed frequency of the light, fo = 641 x 10¹⁴ Hz
Find:
• The frequency of the source, fs
To solve this problem, we have to apply the Doppler Effect formula for light,
[tex]f_o=f_s\sqrt{\frac{1-v_s/c}{1+v_s/c}}[/tex]Solving for fs,
[tex]f_s=f_o\sqrt{\frac{1+v_s/c}{1-v_s/c}}[/tex]Replace the known values and solve,
[tex]f_s=641\cdot10^{14}Hz\cdot\sqrt{\frac{1+(236km/s)/(300,000km/s)}{1-(236km/s)/(300,000km/s)}}\approx6.4\cdot10^{16}Hz[/tex]Hence, the emitted frequency of the comet's light is 6.4 x 10¹⁶ Hz.
