Skip to content
Additional Exercises · 10.11

Q.The 6563 A˚6563\ \text{Å} Hα_\alpha line emitted by hydrogen in a star is found to be red-shifted by 15 A˚15\ \text{Å}. Estimate the speed with which the star is receding from the Earth.

Sikkim CbseNCERTSubjective· 2mImportance★★★★★est
28% · 13/46 Questions
✓ Free question

Using the Doppler shift relation Δλλ=vc\dfrac{\Delta\lambda}{\lambda} = \dfrac{v}{c}, a redshift of 15 A˚15\ \text{Å} on a 6563 A˚6563\ \text{Å} line gives a recession speed of about 6.86×105 m s−16.86\times10^{5}\ \text{m s}^{-1}.

Step 1: The Doppler shift relation for light

For speeds much smaller than cc, the fractional change in wavelength due to relative radial motion between source and observer is:

Δλλ=vc\frac{\Delta\lambda}{\lambda} = \frac{v}{c}

A redshift (Δλ>0\Delta\lambda > 0, wavelength increases) corresponds to the source receding from the observer.

Step 2: Substitute the given values

λ=6563 A˚,Δλ=15 A˚\lambda = 6563\ \text{Å}, \quad \Delta\lambda = 15\ \text{Å}

v=c×Δλλ=3×108 m s−1×156563v = c \times \frac{\Delta\lambda}{\lambda} = 3\times10^{8}\ \text{m s}^{-1} \times \frac{15}{6563}

Step 3: Evaluate

156563≈2.286×10−3\frac{15}{6563} \approx 2.286\times10^{-3}

v≈3×108×2.286×10−3≈6.86×105 m s−1v \approx 3\times10^{8} \times 2.286\times10^{-3} \approx 6.86\times10^{5}\ \text{m s}^{-1}

Since the line is red-shifted (wavelength increased), the star is receding from the Earth.

✓Final answer

v≈6.86×105 m s−1 (≈686 km/s), star receding\boxed{v \approx 6.86\times10^{5}\ \text{m s}^{-1}\ (\approx 686\ \text{km/s}),\ \text{star receding}}

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.