Chapter 16.2, Problem 17PP

To determine

To Calculate: The frequency of light when it is observed from that hydrogen atom by an astronomer on the earth.

Answer to Problem 17PP

The observed frequency is $\boxed{f_{obs}=6.03\times {10}^{14}\text{Hz}}$ .

Explanation of Solution

Given:

The speed of the hydrogen atom is $v=6.55\times {10}^6\text{m/s}$ .

Frequency emitted by hydrogen atom is $f=6.16\times {10}^{14}\text{Hz}$ .

Formula used:

The expression for observed frequency is

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

Here, $f_{obs}$ , $f$ , $v$ and $c$ are observed frequency, actual frequency, velocity of the object and speed of light respectively.

Calculation:

Since the galaxy and the astronomer are moving from each other, therefore negative sign should be selected.

Equation (1) becomes,

  $f_{obs}=f\left(1-\frac{v}{c}\right)$.........(2)

Substitute $6.65\times {10}^6\text{m/s}$ for $v$ and $6.16\times {10}^{14}\text{Hz}$ for $f$ and $3\times {10}^8{\text{ms}}^{\text{-1}}$ in equation (2)

  $\begin{array}{l}{f}_{obs}=6.16\times {10}^{14}\text{Hz}\left(1-\frac{6.55\times {10}^6\text{m}\cdot {\text{s}}^{\text{-1}}}{3\times {10}^8\text{m}\cdot {\text{s}}^{\text{-1}}}\right)\\ \boxed{f_{obs}=6.03\times {10}^{14}\text{Hz}}\end{array}$

Conclusion:

Hence, the observed frequency is $\boxed{f_{obs}=6.03\times {10}^{14}\text{Hz}}$ .