Chapter 16, Problem 6P

(a)

To determine

Wavelength of light emitted from a galaxy lies at $2\times {10}^6\text{lightyears}$ from the earth.

Answer to Problem 6P

Wavelength is $590.2\text{nm}$.

Explanation of Solution

Write the Hubble’s law equation.

    $v=H_{0}R$        (I)

Here, $v$ is the speed of galaxy, $H_0$ is the Hubble’s constant and $R$ is the distance from earth.

Write the relativistic doppler formula.

  ${\lambda }^{\prime }=\frac{\sqrt{1+v/c}}{1-v/c}\lambda$        (II)

Here, ${\lambda }^{\prime }$ is the observed wavelength and $\lambda$ is the original wavelength.

Conclusion:

Substitute $\left(2.3\times {10}^{-2}\text{m}/\text{s}\right)/\text{ly}$ for $H_0$ and $2\times {10}^6\text{lightyears}$ for $R$ in equation (I).

    $\begin{array}{c}v=\left(\left(2.3\times {10}^{-2}\text{m}/\text{s}\right)/\text{ly}\right)\left(2\times {10}^6\text{lightyears}\right)\\ =4.6\times {10}^4\text{m}/\text{s}\end{array}$

Substitute 590 nm for $\lambda$, $4.6\times {10}^4\text{m}/\text{s}$ for $v$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (II).

    $\begin{array}{c}{\lambda }^{\prime }=\sqrt{\frac{1+\left[\left(4.6\times {10}^8\text{m}/\text{s}\right)/\left(3\times {10}^8\text{m}/\text{s}\right)\right]}{1-\left[\left(4.6\times {10}^8\text{m}/\text{s}\right)/\left(3\times {10}^8\text{m}/\text{s}\right)\right]}}\left(590\text{nm}\right)\\ =\left(1.00031\right)\left(590\text{nm}\right)\\ =590.2\text{nm}\end{array}$

Thus, the wavelength is $590.2\text{nm}$.

(b)

To determine

Wavelength of light emitted from a galaxy lies at $2\times {10}^8\text{lightyears}$ from the earth.

Answer to Problem 6P

Wavelength is $599.1\text{nm}$.

Explanation of Solution

Substitute $\left(2.3\times {10}^{-2}\text{m}/\text{s}\right)/\text{ly}$ for $H_0$ and $2\times {10}^8\text{lightyears}$ for $R$ in equation (I).

    $\begin{array}{c}v=\left(\left(2.3\times {10}^{-2}\text{m}/\text{s}\right)/\text{ly}\right)\left(2\times {10}^8\text{lightyears}\right)\\ =4.6\times {10}^6\text{m}/\text{s}\end{array}$

Substitute 590 nm for $\lambda$, $4.6\times {10}^6\text{m}/\text{s}$ for $v$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (II).

    $\begin{array}{c}{\lambda }^{\prime }=\sqrt{\frac{1+\left[\left(4.6\times {10}^8\text{m}/\text{s}\right)/\left(3\times {10}^8\text{m}/\text{s}\right)\right]}{1-\left[\left(4.6\times {10}^8\text{m}/\text{s}\right)/\left(3\times {10}^8\text{m}/\text{s}\right)\right]}}\left(590\text{nm}\right)\\ =\left(1.015\right)\left(590\text{nm}\right)\\ =599.1\text{nm}\end{array}$

Conclusion:

Thus, the wavelength is $599.1\text{nm}$.

(b)

To determine

Wavelength of light emitted from a galaxy lies at $2\times {10}^9\text{lightyears}$ from the earth.

Answer to Problem 6P

Wavelength is $688.6\text{nm}$.

Explanation of Solution

Substitute $\left(2.3\times {10}^{-2}\text{m}/\text{s}\right)/\text{ly}$ for $H_0$ and $2\times {10}^9\text{lightyears}$ for $R$ in equation (I).

    $\begin{array}{c}v=\left(\left(2.3\times {10}^{-2}\text{m}/\text{s}\right)/\text{ly}\right)\left(2\times {10}^9\text{lightyears}\right)\\ =4.6\times {10}^7\text{m}/\text{s}\end{array}$

Substitute 590 nm for $\lambda$, $4.6\times {10}^7\text{m}/\text{s}$ for $v$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (II).

    $\begin{array}{c}{\lambda }^{\prime }=\sqrt{\frac{1+\left[\left(4.6\times {10}^9\text{m}/\text{s}\right)/\left(3\times {10}^8\text{m}/\text{s}\right)\right]}{1-\left[\left(4.6\times {10}^9\text{m}/\text{s}\right)/\left(3\times {10}^8\text{m}/\text{s}\right)\right]}}\left(590\text{nm}\right)\\ =\left(1.16\right)\left(590\text{nm}\right)\\ =688.6\text{nm}\end{array}$

Conclusion:

Thus, the wavelength is $688.6\text{nm}$.