The periodic time of rotation of a certain star is 22 days and its radius is `7 xx 10^(8)` metres . If the wavelength of light emitted by its surface be `4320 Å`, the Doppler shift will be (1 day = 86400 sec )

Options

A. 0.33Å
B. 0.33Å
C. 3.3Å
D. 33Å

Answer

A

Explanation

The Doppler shift `Delta lambda ` is given by
`(Delta lambda)/(lambda)=(v)/(c) "but" v= r omega=r* (2pi)/(T)`
`:. (Delta lamdba)/(lambda)=(2pi r)/(T xx c)`
`:. Delta lambda=(2pi r * lambda)/(T xx c)`
`=(2 xx 3.14 xx 7 xx 10^(8) xx 4320 xx 10^(-10))/(22 xx 86400 xx 3 xx 10^8)`
`=(7 xx 3.14 xx 10^(-10))/( 11 xx 20 xx 3)`
`=0.033 xx 10^(-10) m =0.033 Å`