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The periodic time of rotation of a certain star is 22 days and its radius is 7×108 meters. If the wavelength of light emitted by its surface be 4320 A0, the Doppler shift will be (1 day = 86400 sec)   

a
0.033A0
b
0.33A0
c
3.3A0
d
33A0

detailed solution

Correct option is A

ω=2πT=2×227×22×86400 v=rω     =7×1082×227×22×86400=106432 ∆λ=vCλ=1063×108×1432×4320=0.033A0

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