A star, which is emitting radiation at a wavelength of 5000 $\mathrm{\AA }$, is approaching the earth with a velocity of 1.50 × 10⁶ m/s. The change in wavelength of the radiation as received on the earth is

Given : Wavelength (λ) = 5000 $\mathrm{\AA }$ 

velocity of star (v) = 1.5 × 10⁶ m/s. 

We know that wavelength of the approaching star (λ’) $=\lambda \frac{c-v}{c}$

or, $\frac{\lambda \text{'}}{\lambda }=\frac{c-v}{c}=1-\frac{v}{c}$

$\mathrm{or}, \frac{v}{c}=1-\frac{\lambda \text{'}}{\lambda }=\frac{\lambda -\lambda \text{'}}{\lambda }=\frac{∆\lambda }{\lambda }.$Therefore,

     $∆\lambda =\lambda \times \frac{v}{c}=5000\times \frac{1.5\times {10}^6}{3\times {10}^8}=25$ $\mathrm{\AA }$.

   [where ∆λ = Change in the wavelength]