Chapter 16, Problem 73A

To determine

The wavelength of hydrogen observed from the receding edge and the approaching edge.

Answer to Problem 73A

The wavelength of hydrogen observed from the receding edge and the approaching edge is $\Delta \lambda =\pm 3.3\times {10}^{-12}\text{ m}$ .

Explanation of Solution

Given:

The period of rotation is $t=25\text{ days}$ .

The diameter of the sun is $d=1.4\times {10}^9\text{ m}$

The frequency of light is $f=6.16\times {10}^{14}\text{ Hz}$

Formula used:

The expression for the speed of rotation is,

$v=\frac{\pi d}{t}$

Here, $d$ is the diameter and $t$ is the time period.

The expression for the wavelength is,

$\lambda =\frac{c}{f}$

Here, $c$ is the speed of the light and $f$ is the frequency.

The expression for the change in wavelength is,

$\Delta \lambda =\pm \frac{v}{c}\lambda$

Here, $v$ is the speed of rotation and $\lambda$ is the wavelength.

Calculation:

Consider the speed of light as $c=3.0\times {10}^8\text{m}/\text{s}$ .

The speed of rotation is

$\begin{array}{l}{v}_{\text{rot}}=\frac{\pi d}{t}\\ v_{\text{rot}}=\frac{\pi \times \left(1.4\times {10}^9\text{ m}\right)}{\left(25\text{ days}\times \frac{24\text{ hrs}}{1\text{ day}}\times \frac{60\text{ mins}}{1\text{ hr}}\times \frac{60\text{ s}}{1\text{ min}}\right)}\\ v_{\text{rot}}=2.04\times {10}^3\text{m}/\text{s}\end{array}$

The wavelength of the light is

$\begin{array}{l}\lambda =\frac{c}{f}\\ \lambda =\frac{\left(3.0\times {10}^8\text{m}/\text{s}\right)}{\left(6.16\times {10}^{14}\text{ Hz}\right)}\\ \lambda =4.9\times {10}^{-7}\text{ m}\end{array}$

The wavelength of hydrogen observed from the receding edge and the approaching edge is,

$\begin{array}{l}\Delta \lambda =\pm \frac{v_{\text{rot}}}{c}\lambda \\ \Delta \lambda =\pm \frac{\left(2.04\times {10}^3\text{m}/\text{s}\right)}{\left(3.0\times {10}^8\text{m}/\text{s}\right)}\left(4.9\times {10}^{-7}\text{ m}\right)\\ \Delta \lambda =\pm 3.3\times {10}^{-12}\text{ m}\end{array}$

Conclusion:

Thus, the wavelength of hydrogen observed from the receding edge and the approaching edge is $\Delta \lambda =\pm 3.3\times {10}^{-12}\text{ m}$ .