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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 )A. 0.33ÅB. 0.33ÅC. 3.3ÅD. 33Å |
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Answer» Correct Answer - A 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` |
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