Question

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

Answer 1

Step 1: Given information

The wavelength, $\mathrm{\lambda }=5000\mathrm{A}$

Velocity, $\mathrm{v}=1.5\mathrm{x}{10}^4\mathrm{m}/\mathrm{s}$

Step 2: We have to find the change in wavelength of the radiation as received on the earth.

Step 3: Calculation

The wavelength of the approaching star,

$$\begin{gathered} \frac{\mathrm{\lambda }\text{'}}{\mathrm{\lambda }}=1–\left(\frac{\mathrm{v}}{\mathrm{c}}\right) \\ \frac{\mathrm{v}}{\mathrm{c}}=1–\left(\frac{\mathrm{\lambda }\text{'}}{\mathrm{\lambda }}\right) \\ =\frac{\mathrm{\lambda }–\mathrm{\lambda }\text{'}}{\mathrm{\lambda }} \\ =\frac{∆\mathrm{\lambda }}{\mathrm{\lambda }} \end{gathered}$$

The change in wavelength be:

$\mathrm{\Delta \lambda }=\mathrm{\lambda }\times \left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

By substituting the values,

$$\begin{gathered} =5000\times \left[\frac{(1.5\times 106)}{3}\times {10}^8\right] \\ =25\mathrm{A} \end{gathered}$$

Therefore, the change in wavelength is $25\mathrm{A}$.