Question

A star, emitting a radiation of wavelength $5000\mathring{\text{A}}$, is approaching earth with a speed of $3\times {10}^6\text{m/s}$. The wavelength of the radiation received on earth will be,

• A. $5100\mathring{\text{A}}$
• B. $4900\mathring{\text{A}}$
• C. $5050\mathring{\text{A}}$
• D. $4950\mathring{\text{A}}$

Answer 1

The correct option is D $4950\mathring{\text{A}}$

The Doppler shift in wavelength is given by,

$\frac{\Delta \lambda }{\lambda }=\frac{v_r}{c}\Rightarrow \Delta \lambda =\lambda \times \frac{v_r}{c}$

Here,
$\lambda =5\times {10}^{-7}\text{m},c=3\times {10}^8\text{m/s}$
$v_r=3\times {10}^6\text{m/s}$

$\Delta \lambda =\frac{3\times {10}^6}{3\times {10}^8}\times 5\times {10}^{-7}=5\times {10}^{-9}\text{m}$

$\therefore \Delta \lambda =50\mathring{\text{A}}$

As the star is approaching earth, the apparent wavelength will be less than that of actual wavelength.

$\therefore$ wavelength of its radiation received on earth is,
$=\lambda -\Delta \lambda =5000-50=4950\mathring{\text{A}}$

Hence, $\left(D\right)$ is the correct answer.