Question

The earth is moving towards a fixed star with a velocity of 30 km s^-1 . An observer on the earth observes a shift of 0.58 Å in the wavelength of light coming from the star. The actual wavelength of light emitted by the star:

• A. 5800 Å
• B. 2400 Å
• C. 1200 Å
• D. 6000 Å

Answer 1

Answer: (A)

5800 Å

Answer 2

: using $\Delta\lambda = \frac{v}{c}\lambda$

Here $v = 30kms^{- 1} = 30 \times 10^{3}ms^{- 1}$

$c = 3 \times 10^{8}ms^{- 1}\Delta\lambda = 0.58Å$

$\therefore\lambda = \frac{c}{v}.\Delta\lambda = \frac{3 \times 10^{8}}{30 \times 10^{3}} \times 0.58 = 5800Å$