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Q.

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)

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a

0.033A0

b

0.33A0

c

3.3A0

d

33A0

answer is A.

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Detailed Solution

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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