Chapter 31, Problem 41P

(a)

To determine

The wavelength of the sodium line emitted from galaxies.

Answer to Problem 41P

The wavelength of the sodium line emitted from galaxies is $\underset{\_}{590.09\text{nm}}$.

Explanation of Solution

Write the expression for velocity from Hubble’s law.

$v=HR$        (I)

Here, $v$ is the velocity, $H$ is the Hubble’s constant and $R$ is the distance.

Write the expression for the wavelength by using equation (I).

$\begin{array}{c}\lambda \text{'}=\lambda \sqrt{\frac{1+\left(v/c\right)}{1-\left(v/c\right)}}\\ =\lambda \sqrt{\frac{1+\left(HR/c\right)}{1-\left(HR/c\right)}}\end{array}$        (II)

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

Conclusion:

Substitute, $22\times {10}^{-3}\text{m/s}\text{.ly}$ for $H$, $2.00\times {10}^6\text{ly}$ for $R$, $2.998\times {10}^8\text{m/s}$ for $c$, $590\text{nm}$ for $\lambda$ in equation (II) to find ${\lambda }^{\prime }$.

$\begin{array}{c}\lambda \text{'}=\left(590\text{nm}\right)\sqrt{\frac{1+\left(\left(22\times {10}^{-3}\text{m/s}\text{.ly}\right)\left(2.00\times {10}^6\text{ly}\right)/\left(2.998\times {10}^8\text{m/s}\right)\right)}{1-\left(\left(22\times {10}^{-3}\text{m/s}\text{.ly}\right)\left(2.00\times {10}^6\text{ly}\right)/\left(2.998\times {10}^8\text{m/s}\right)\right)}}\\ =590.09\text{nm}\end{array}$

Thus, the wavelength of the sodium line emitted from galaxies is $\underset{\_}{590.09\text{nm}}$.

(b)

To determine

The wavelength of the sodium line emitted from galaxies.

Answer to Problem 41P

The wavelength of the sodium line emitted from galaxies is $\underset{\_}{599\text{nm}}$.

Explanation of Solution

Write the expression for velocity from Hubble’s law.

$v=HR$        (I)

Here, $v$ is the velocity, $H$ is the Hubble’s constant and $R$ is the distance.

Write the expression for the wavelength by using equation (I).

$\begin{array}{c}\lambda \text{'}=\lambda \sqrt{\frac{1+\left(v/c\right)}{1-\left(v/c\right)}}\\ =\lambda \sqrt{\frac{1+\left(HR/c\right)}{1-\left(HR/c\right)}}\end{array}$        (II)

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

Conclusion:

Substitute, $22\times {10}^{-3}\text{m/s}\text{.ly}$ for $H$, $2.00\times {10}^8\text{ly}$ for $R$, $2.998\times {10}^8\text{m/s}$ for $c$, $590\text{nm}$ for $\lambda$ in equation (II) to find ${\lambda }^{\prime }$.

$\begin{array}{c}\lambda \text{'}=\left(590\text{nm}\right)\sqrt{\frac{1+\left(\left(22\times {10}^{-3}\text{m/s}\text{.ly}\right)\left(2.00\times {10}^8\text{ly}\right)/\left(2.998\times {10}^8\text{m/s}\right)\right)}{1-\left(\left(22\times {10}^{-3}\text{m/s}\text{.ly}\right)\left(2.00\times {10}^8\text{ly}\right)/\left(2.998\times {10}^8\text{m/s}\right)\right)}}\\ =599\text{nm}\end{array}$

Thus, the wavelength of the sodium line emitted from galaxies is $\underset{\_}{599\text{nm}}$.

(c)

To determine

The wavelength of the sodium line emitted from galaxies.

Answer to Problem 41P

The wavelength of the sodium line emitted from galaxies is $\underset{\_}{684\text{nm}}$.

Explanation of Solution

Write the expression for velocity from Hubble’s law.

$v=HR$        (I)

Here, $v$ is the velocity, $H$ is the Hubble’s constant and $R$ is the distance.

Write the expression for the wavelength by using equation (I).

$\begin{array}{c}\lambda \text{'}=\lambda \sqrt{\frac{1+\left(v/c\right)}{1-\left(v/c\right)}}\\ =\lambda \sqrt{\frac{1+\left(HR/c\right)}{1-\left(HR/c\right)}}\end{array}$        (II)

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

Conclusion:

Substitute, $22\times {10}^{-3}\text{m/s}\text{.ly}$ for $H$, $2.00\times {10}^9\text{ly}$ for $R$, $2.998\times {10}^8\text{m/s}$ for $c$, $590\text{nm}$ for $\lambda$ in equation (II) to find ${\lambda }^{\prime }$.

$\begin{array}{c}\lambda \text{'}=\left(590\text{nm}\right)\sqrt{\frac{1+\left(\left(22\times {10}^{-3}\text{m/s}\text{.ly}\right)\left(2.00\times {10}^9\text{ly}\right)/\left(2.998\times {10}^8\text{m/s}\right)\right)}{1-\left(\left(22\times {10}^{-3}\text{m/s}\text{.ly}\right)\left(2.00\times {10}^9\text{ly}\right)/\left(2.998\times {10}^8\text{m/s}\right)\right)}}\\ =684\text{nm}\end{array}$

Thus, the wavelength of the sodium line emitted from galaxies is $\underset{\_}{684\text{nm}}$.