A star emitting light of wavelength 5896 mat | vedclass

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(D) The Doppler shift in wavelength for a source moving away from the observer is given by $\Delta \lambda = \frac{v}{c} \lambda$.Given: $\lambda = 58

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Increase by $70.75 \ \mathring{A}$

Vedclass · Answer

(D) The Doppler shift in wavelength for a source moving away from the observer is given by $\Delta \lambda = \frac{v}{c} \lambda$.Given: $\lambda = 5896 \ \mathring{A}$,$v = 3600 \ km/s = 3.6 	imes 10^6 \ m/s$,and $c = 3 	imes 10^8 \ m/s$.Substituting the values: $\Delta \lambda = \frac{3.6 	imes 10^6}{3 	imes 10^8} 	imes 5896$.$\Delta \lambda = 1.2 	imes 10^{-2} 	imes 5896 = 70.752 \ \mathring{A}$.Since the star is moving away,the observed wavelength increases. Thus,the wavelength increases by $70.75 \ \mathring{A}$.

$A$ star emitting light of wavelength $5896 \ \mathring{A}$ is moving away from the earth with a speed of $3600 \ km/s$. The wavelength of light observed on earth will be:
($c = 3 \times 10^8 \ m/s$ is the speed of light)

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$A$ star emitting light of wavelength $5896 \ \mathring{A}$ is moving away from the earth with a speed of $3600 \ km/s$. The wavelength of light observed on earth will be:($c = 3 \times 10^8 \ m/s$ is the speed of light)