The 6563 mathring{A} line emitted by a hydro | vedclass

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(D) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the shift in wavelen

Vedclass · Accepted Answer

$2.29 	imes 10^{5} \ m/s$

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(D) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the shift in wavelength,$\lambda$ is the original wavelength,$v$ is the velocity of the source,and $c$ is the speed of light.Given: $\lambda = 6563 \ \mathring{A}$,$\Delta \lambda = 5 \ \mathring{A}$,and $c = 3 	imes 10^{8} \ m/s$.Rearranging the formula to solve for $v$: $v = \frac{\Delta \lambda}{\lambda} 	imes c$.Substituting the values: $v = \frac{5}{6563} 	imes 3 	imes 10^{8} \ m/s$.$v \approx 0.0007618 	imes 3 	imes 10^{8} \ m/s$.$v \approx 2.2855 	imes 10^{5} \ m/s \approx 2.29 	imes 10^{5} \ m/s$.

The $6563 \ \mathring{A}$ line emitted by a hydrogen atom in a star is found to be red-shifted by $5 \ \mathring{A}$. The speed with which the star is receding from the Earth is:

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$A$ spectral line $\lambda = 5000\ \mathring A$ in the light coming from a distant star is observed as $5200\ \mathring A$. What will be the recession velocity of the star?

If a star is moving towards the Earth,then the spectral lines are shifted towards:

The wavelength of light coming from a star is $3700 \, \mathring{A}$. If the observed wavelength from Earth is $3737 \, \mathring{A}$,what is the velocity of the star? [$c = 3 \times 10^8 \, m/s$]

If the shift of wavelength of light emitted by a star is towards the violet end of the spectrum,then this shows that the star is

Light coming from a star is observed to have a wavelength of $3737 \ \mathring{A}$,while its real wavelength is $3700 \ \mathring{A}$. The speed of the star relative to the earth is (Speed of light $c = 3 \times 10^8 \ m/s$).

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