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(B) The formula for the Doppler shift in wavelength for a source moving away from the observer is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(B) The formula for the Doppler shift in wavelength for a source moving away from the observer is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$.Here,the speed of the rocket $v = 10^{6} \ m/s$,the speed of light $c = 3 	imes 10^{8} \ m/s$,and the original wavelength $\lambda = 5700 \ \mathring{A}$.Substituting these values into the formula:$\Delta \lambda = \lambda 	imes \frac{v}{c}$$\Delta \lambda = 5700 \ \mathring{A} 	imes \frac{10^{6} \ m/s}{3 	imes 10^{8} \ m/s}$$\Delta \lambda = 5700 	imes \frac{1}{300} \ \mathring{A}$$\Delta \lambda = 19 \ \mathring{A}$.

$A$ rocket is moving away from the Earth at a speed of $10^{6} \ m/s$. If the wavelength of the light wave emitted by it is $5700 \ \mathring{A}$,what will be its Doppler shift in $\mathring{A}$?

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