Chapter 17, Problem 36Q

(a)

To determine

The graph for the intensity curve of a blackbody having a temperature of $3000\text{K}$ with reference to the figure given below.

Answer to Problem 36Q

Solution:

Explanation of Solution

Introduction:

Wien’s law: Objects at different temperatures emit a spectrum that peaks at different wavelengths.

Explanation:

According to Wien’s law, wavelength is inversely proportional to temperature, so objects at different temperatures emit a spectrum that peaks at different wavelengths.

Now, as the wavelengths of the spectrum are different, the colors associated with those wavelengths will be different as well.

The intensity curve is plotted by representing wavelength on the horizontal axis and intensity on the vertical axis. So, for a temperature of $3000\text{K}$, the intensity curve can be sketched as,

The curve will peak at around $970\text{nm}$ for a black body of temperature $3000\text{K}$.

The blackbody will appear red in color because the intensity of the radiation emitting from the blackbody is peaking at a longer wavelength. The human eye can sense red, green, and blue colors.

According to the above graph, the maximum intensity is for the red color, so the blackbody will appear red in color to us.

Conclusion:

The sketch shows that for a temperature of $3000\text{K}$, which indicates a relatively cold star, the intensity peaks at longer wavelengths of about $970\text{nm}$. So, the star will appear red in color.

(b)

To determine

The graph for the intensity curve of a blackbody having a temperature of $12000\text{K}$ with reference to the figure given below.

Answer to Problem 36Q

Solution:

Explanation of Solution

Introduction:

Wien’s law: Objects at different temperatures emit a spectrum that peaks at different wavelengths because the wavelength is inversely proportional to the temperature.

Explanation:

According to Wien’s law, objects at different temperatures emit a spectrum that peaks at different wavelengths.

As the wavelengths of the spectrum are different, the colors associated with those wavelengths will be different as well.

The intensity curve is plotted by representing wavelength on the horizontal axis and intensity on the vertical axis.

So, for a temperature of $12000\text{K}$, the intensity curve can be sketched as,

The curve will peak at around $240\text{nm}$ for a blackbody of temperature $12000\text{K}$.

The black body will appear blue in color because the intensity of the radiation emitting from the blackbody is peaking at a shorter wavelength. The human eye can sense red, green, and blue colors.

According to the graph above, the maximum intensity is for the blue color, so the blackbody will appear blue in color to us.

Conclusion:

The sketch shows that for a temperature of $12000\text{K}$, which indicates a relatively hot star, the intensity peaks at shorter wavelengths of about $240\text{nm}$. So, the star will appear blue in color to us.

(c)

To determine

To explain: The reason behind color ratios $\frac{b_{\text{V}}}{b_{\text{B}}}$ & $\frac{b_{\text{B}}}{b_{\text{U}}}$ being less than 1 for very hot stars, and greater than 1 for very cool stars, with reference to the results obtained in parts (a) and (b).

Answer to Problem 36Q

Solution:

For a relatively hot star, the value of brightness is in order $b_{\text{U}}>b_B>b_{\text{V}}$ and for a cool star, it is $b_{\text{V}}>b_{\text{B}}>b_{\text{U}}$.

So, the ratios $\frac{b_{\text{V}}}{b_{\text{B}}}$ & $\frac{b_{\text{B}}}{b_{\text{U}}}$ will be greater than 1 for a cold star and less than 1 for a hot star.

Explanation of Solution

Introduction:

When a hot star is observed through the U (ultraviolet), V (yellow and green), and B (blue) filters, the intensity of the U filter will be higher than that of the B filter, and will be the dimmest when observed through the V filter.

Similarly, when a cold star is observed through U, V, and B filters, the intensity will be higher through the V filter than through the B filter, and it will be dimmest through U filter.

Explanation:

For a star of temperature $3000\text{ K}$, which indicates a relatively cold star, the value of brightness in decreasing order will be, $b_{\text{V}}>b_{\text{B}}>b_{\text{U}}$.

So, the value of the ratios $\frac{b_{\text{V}}}{b_{\text{B}}}$ & $\frac{b_{\text{B}}}{b_{\text{U}}}$ will always be greater than 1.

For a star of temperature $12000\text{ K}$, which indicates a relatively hot star, the value of brightness in decreasing order will be, $b_{\text{U}}>b_B>b_{\text{V}}$.

So, the value of the ratios $\frac{b_{\text{V}}}{b_{\text{B}}}$ & $\frac{b_{\text{B}}}{b_{\text{U}}}$ will always be less than 1.

Conclusion:

Hot stars appear to be blue in color and cold stars appear to be red.

This is why when a hot star is observed through a B filter, the intensity is highest and similarly, when a cold star is observed through a V filter, the intensity is highest.