Chapter 16, Problem 56Q

To determine

The wavelengths of radiation at which the photosphere, chromosphere, and corona emits the most radiation and, using these wavelengths, mention the best way for observing such regions in the solar atmosphere, given that each part of the atmosphere behaves as a perfect blackbody and average temperature for the chromosphere is 50000 K and that for corona is $1.5\times {10}^6\text{K}$.

Answer to Problem 56Q

Solution:

The wavelengths at which the photosphere, chromosphere, and corona emits the most radiation are $500\text{nm}$, $58\text{nm}$, and $2\text{nm}$, respectively.

This result shows that the wavelength of light from photosphere is much larger. So, the observation of photosphere can be possible from Earth, whereas the observation of chromosphere and corona is possible from space station.

Explanation of Solution

Given data:

The temperature of chromosphere is $50000\text{K}$.

The temperature of corona is $\text{1}\text{.5}\times {\text{10}}^{\text{6}}\text{K}$.

Formula used:

Wien’s law is written as,

${\lambda }_{\max }\text{=}\frac{b}{T}$

Here, ${\lambda }_{\max }$ is the maximum wavelength, T is the absolute temperature in kelvin, and b is the constant of proportionality which is equal to $2.89\times {10}^{-3}\text{m}\cdot \text{K}$.

Explanation:

Consider the average surface temperature of photosphere to be 5800 K.

Use Wien’s law for writing the expression for wavelength emitted by the photosphere.

${\lambda }_{\max ,1}\text{=}\frac{b}{T_1}$

Here, subscript 1 represents the corresponding quantities for photosphere.

Substitute $2.89\times {10}^{-3}\text{m}\cdot \text{K}$ for b and $5800\text{K}$ for $T_1$.

$\begin{array}{c}{\lambda }_{\max ,1}\text{=}\frac{2.89\times {10}^{-3}\text{m}\cdot \text{K}}{5800\text{K}}\\ =4.98\times {10}^{-7}\text{ m}\left(\frac{{10}^9\text{ nm}}{1\text{ m}}\right)\\ \simeq 500\text{ nm}\end{array}$

Since the wavelength of $500\text{nm}$ lies in the range of visible radiation, the photosphere emits visible light radiation.

Use Wien’s law for writing the expression for wavelength emitted by the chromosphere.

${\lambda }_{\max ,2}\text{=}\frac{b}{T_2}$

Here, subscript 2 represents the corresponding quantities for chromosphere.

Substitute $2.89\times {10}^{-3}\text{m}\cdot \text{K}$ for b and $50000\text{K}$ for $T_2$.

$\begin{array}{c}{\lambda }_{\max ,2}\text{=}\frac{2.89\times {10}^{-3}\text{m}\cdot \text{K}}{50000\text{K}}\\ =5.78\times {10}^{-8}\text{ m}\left(\frac{{10}^9\text{ nm}}{1\text{ m}}\right)\\ \simeq 58\text{ nm}\end{array}$

Since the wavelength of $58\text{nm}$ lies in the range of ultraviolet radiation, the chromosphere emits ultraviolet radiation.

Use Wien’s law for writing the expression for wavelength emitted by the corona.

${\lambda }_{\max ,3}\text{=}\frac{b}{T_3}$

Here, subscript 2 represents the corresponding quantities for corona.

Substitute $2.89\times {10}^{-3}\text{m}\cdot \text{K}$ for b and $\text{1}\text{.5}\times {\text{10}}^{\text{6}}\text{K}$ for $T_3$.

$\begin{array}{c}{\lambda }_{\max ,3}\text{=}\frac{2.89\times {10}^{-3}\text{m}\cdot \text{K}}{\text{1}\text{.5}\times {\text{10}}^{\text{6}}\text{K}}\\ =1.93\times {10}^{-9}\text{ m}\left(\frac{{10}^9\text{ nm}}{1\text{ m}}\right)\\ \simeq 2\text{ nm}\end{array}$

Since the wavelength of $2\text{nm}$ lies in the range of x-ray radiation, the corona emits x-ray radiation.

Chromosphere and corona emit light of short wavelength, which is not clearly seen by the observer. For getting a clear picture of corona and chromosphere, space station is used.

Conclusion:

Hence, the wavelengths at which the photosphere, chromosphere, and corona emit the most radiation are $500\text{nm}$, $\text{58}\text{nm}$, and $2\text{nm}$, respectively.