Question

A star, which is emitting radiation at a wavelength of $5000$ $�$ is approaching the earth with a velocity of $1.5 \times 10^4m/s.$ The change in wavelength of the radiation as received on the earth is

• A. $25$ $�$
• B. $100$ $�$
• C. zero
• D. $2.5$ $�$

Answer 1

Answer: (A)

$25$ $�$

Answer 2

Wavelength $(\lambda)=5000$ $�$ and velocity $(v)=1.5\times10^4 m/s.$
Wavelength of the approaching star, $(\lambda')=\lambda\frac{c-v_s}{c}$ or $\frac{\lambda'}{\lambda}=1-\frac{v}{c}$ or, $\frac{v}{c}=1-\frac{\lambda'}{\lambda}=\frac{\lambda-\lambda'}{\lambda}=\frac{\Delta\, \lambda}{\lambda}$
Therefore $\Delta \lambda=\lambda \times\frac{\lambda}{c}=5000 \,�\times\frac{1.5\times 10^6}{3\times10^8}=25$ $�$.
(where $\Delta \lambda$ is the change in the wavelength)