Chapter 16, Problem 2STP

To determine

To Choose: The correct option.

Answer to Problem 2STP

C. $5.7\times {10}^{14}\text{Hz}$

Explanation of Solution

Given:

The galaxy is moving away with a speed, $v=5.8\times {10}^6\text{}\text{m/s}$

The observed frequency of the light, $f_{\text{obs}}=5.6\times {10}^{14}\text{Hz}$

Formula used:

Doppler effect: The observed frequency of light can be written in terms of actual frequency of light $\left(f\right)$ and the relative velocity of the source with respect to the observer $\left(v\right)$ as,

  $f_{\text{obs}}=f\left(1\pm \frac{v}{c}\right)$

Where $c$ is the speed of light.

Now, if the observer and the source are approaching each other, then the observed frequency is,

  $f_{\text{obs}}=f\left(1+\frac{v}{c}\right)$

And, if the observer and the source are receding away from each other, then the observed frequency is,

  $f_{\text{obs}}=f\left(1-\frac{v}{c}\right)$

Calculation:

The speed of light is known as, $c=3\times {10}^8\text{}\text{m/s}$

Thus for the given case, using the given values in above formula,

  $\begin{array}{c}{f}_{\text{obs}}=f\left(1-\frac{v}{c}\right)\\ 5.6\times {10}^{14}=f\left(1-\frac{5.8\times {10}^6}{3.0\times {10}^8}\right)\\ f\approx 5.7\times {10}^{14}\text{Hz}\end{array}$

Conclusion:

Hence, the emitted frequency of the light is $5.7\times {10}^{14}\text{Hz}$ .