Chapter 9, Problem 9.26P

(a)

To determine

The number of absorbing sodium atoms per unit area of the photosphere.

Answer to Problem 9.26P

The number of absorbing sodium atoms per unit area of the photosphere ${10}^{19}{N}_a$.

Explanation of Solution

Write the expression for the calculation of ${\log }_{10}\left(\frac{W}{\lambda }\right)$.

  ${\log }_{10}\left(\frac{W}{\lambda }\right)=0$        (1)

Write the expression for general curve of the growth for the sun.

  ${\log }_{10}\left[f{N}_a\left(\frac{\lambda }{500}\right)\text{nm}\right]=0$        (2)

Here, $\lambda$ is the wavelength.

Conclusion:

Substitute $0.0067$ for $W$ and $330.298$ for $\lambda$ in the equation (1),

  $\begin{array}{l}{\log }_{10}\left(\frac{W}{\lambda }\right)={\log }_{10}\left(\frac{0.0067}{330.298}\right)\\ {\log }_{10}\left(\frac{W}{\lambda }\right)=-4.69\end{array}$

Substitute $0.0560$ for $W$ and $589.594$ for $\lambda$ in the equation (1),

  $\begin{array}{l}{\log }_{10}\left(\frac{W}{\lambda }\right)={\log }_{10}\left(\frac{0.0560}{589.594}\right)\\ {\log }_{10}\left(\frac{W}{\lambda }\right)=-4.02\end{array}$

Substitute $330.298\text{nm}$ for $\lambda$ in equation (2),

  ${\log }_{10}\left[f{N}_a\left(\frac{\lambda }{500}\right)\text{nm}\right]=16.63$

Substitute $589.594\text{nm}$ for $\lambda$ in equation (2),

  ${\log }_{10}\left[f{N}_a\left(\frac{\lambda }{500}\right)\text{nm}\right]=18.55$

Substitute $330.298\text{nm}$ for $\lambda$ and $0.0049$ in ${\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\text{nm}\right]$,

  $\begin{array}{l}{\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]={\log }_{10}\left(\frac{0.0049\times 330.298}{500}\right)\\ {\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]=-2.48\end{array}$

Substitute $589.594\text{nm}$ for $\lambda$ and $0.325$ for $f$ in ${\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]$,

  $\begin{array}{l}{\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]={\log }_{10}\left(\frac{0.325\times 589.594}{500}\right)\\ {\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]=-0.41\end{array}$

Calculate the value of the ${\log }_{10}{N}_{\alpha }$ by the difference in the equation (2) and the ${\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]$,

  $\begin{array}{c}{\log }_{10}{N}_a={\log }_{10}\left[\frac{f{N}_a\lambda }{500\text{nm}}\right]-{\log }_{10}\left[\frac{f\lambda }{500\text{nm}}\right]\\ {\log }_{10}{N}_a=16.63+2.48\\ {\log }_{10}{N}_a=19.11\end{array}$

For $\lambda =330.298\text{nm}$

  $\begin{array}{l}{\log }_{10}{N}_a=18.55+0.41\\ {\log }_{10}{N}_a=18.96\end{array}$

For $\lambda =589.594\text{nm}$

As per the two values, the average value of ${\log }_{10}{N}_a=19.04$.

It clearly means that the number of atoms per unit area of the sun’s photosphere is ${10}^{19}{N}_a$.

(b)

To determine

The average value of the number of absorbing sodium atoms per unit area of the photosphere and graph obtained by the average values.

Answer to Problem 9.26P

The average value of the number of absorbing sodium atoms per unit area of the photosphere and graph obtained by the average values. ${\log }_{10}{N}_a=19$.

Explanation of Solution

Write the expression for the calculation of ${\log }_{10}\left(\frac{W}{\lambda }\right)$.

  ${\log }_{10}\left(\frac{W}{\lambda }\right)=0$        (1)

Write the expression for general curve of the growth for the sun.

  ${\log }_{10}\left[f{N}_a\left(\frac{\lambda }{500}\right)\right]=0$        (2)

Here, $\lambda$ is the wavelength.

Conclusion:

Substitute $0.0067$ for $W$ and $330.298$ for $\lambda$ in the equation (1),

  $\begin{array}{l}{\log }_{10}\left(\frac{W}{\lambda }\right)=0={\log }_{10}\left(0.0067/330.298\right)\\ {\log }_{10}\left(\frac{W}{\lambda }\right)=-4.69\end{array}$

Substitute $0.0560$ for $W$ and $589.594$ for $\lambda$ in the equation (1),

  $\begin{array}{l}{\log }_{10}\left(\frac{W}{\lambda }\right)={\log }_{10}\left(0.0560/589.594\right)\\ {\log }_{10}\left(\frac{W}{\lambda }\right)=-4.02\end{array}$

Substitute $330.298\text{nm}$ for $\lambda$ in equation (2),

  ${\log }_{10}\left[f{N}_a\left(\frac{\lambda }{500}\right)\right]=16.63$

Substitute $589.594\text{nm}$ for $\lambda$ in equation (2),

  ${\log }_{10}\left[f{N}_a\left(\frac{\lambda }{500}\right)\right]=18.55$

Substitute $330.298\text{nm}$ for $\lambda$ and $0.0049$ for $f$ in ${\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]$,

  $\begin{array}{l}{\log }_{10}\left(\frac{f}{500}\right)={\log }_{10}\left(\frac{0.0049\times 330.298}{500}\right)\\ {\log }_{10}\left(\frac{f}{500}\right)=-2.48\end{array}$

Substitute $589.594\text{nm}$ for $\lambda$ and $0.325$ for $f$ in ${\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]$

  $\begin{array}{l}{\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]={\log }_{10}\left(\frac{0.325\times 589.594}{500}\right)\\ {\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]=-0.41\end{array}$

Calculate the value of the ${\log }_{10}{N}_{\alpha }$ by the difference in the equation (2) and the ${\log }_{10}\left[f\left(\frac{\lambda }{500}\right)\right]$,

  $\begin{array}{l}{\log }_{10}{N}_a={\log }_{10}\left[\frac{f{N}_a\lambda }{500\text{nm}}\right]-{\log }_{10}\left[\frac{f\lambda }{500\text{nm}}\right]\\ {\log }_{10}{N}_a=16.63+2.48\\ {\log }_{10}{N}_a=19.11\end{array}$

For $\lambda =330.298\text{nm}$

  $\begin{array}{l}{\log }_{10}{N}_a=18.55+0.41\\ {\log }_{10}{N}_a=18.96\end{array}$

For $\lambda =589.594\text{nm}$

As per the two values, the average value of ${\log }_{10}{N}_a=19.04$.

It clearly means that the number of atoms per unit area of the sun’s photosphere is ${10}^{19}{N}_a$.

As per the calculation, by the average of the value of the ${\log }_{10}{N}_a$ as per the problem and the example 9.5.5 will give the value as average value of ${\log }_{10}{N}_a=19$.

With the help of the average values the graph is plot below

Figure (1)