Chapter 16.2, Problem 18PP

To determine

To Calculate: The wavelength that is observed from hydrogen atom.

Answer to Problem 18PP

The observed wavelength is $\boxed{{\lambda }_{obs}=496.61\text{nm}}$ .

Explanation of Solution

Given:

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

Wavelength emitted by hydrogen atom is $\lambda =486\text{nm}$ .

Formula used:

The Doppler shift in the wavelength of light emitted by hydrogen atom moving away from Earth is

  $\left({\lambda }_{obs}-\lambda \right)=\frac{v}{c}\lambda$........(1)

Here, ${\lambda }_{obs}$ is the observed wavelength of hydrogen atom from the Earth, $\lambda$ is the actual wavelength of hydrogen atom in a galaxy, $v$ is the speed of hydrogen atom in a galaxy moving away from earth and $c$ is the velocity of light.

Calculation:

Substitute $6.55\times {10}^6\text{m/s}$ for $v$ , $3\times {10}^8\text{m}\cdot {\text{s}}^{\text{-1}}$ for $c$ and $486\text{nm}$ for $\lambda$ in equation (1)

  $\begin{array}{l}\left({\lambda }_{obs}-\lambda \right)=\frac{v}{c}\lambda \\ {\lambda }_{obs}=\frac{v}{c}\lambda +\lambda \\ {\lambda }_{obs}=\lambda \left(\frac{v}{c}+1\right)\end{array}$

  $\begin{array}{l}{\lambda }_{obs}=486\text{nm}\left(\frac{6.55\times {10}^6\text{m/s}}{3\times {10}^8\text{m/s}}+1\right)\\ {\lambda }_{obs}=486\times {10}^{-9}\times \left(\frac{6.55\times {10}^6\text{m/s}}{3\times {10}^8\text{m/s}}+1\right)\\ \boxed{{\lambda }_{obs}=496.61\text{nm}}\end{array}$

Conclusion:

Hence, the observed wavelength is $\boxed{{\lambda }_{obs}=496.61\text{nm}}$ .