91Ó°ÊÓ

At the surface of the sun, the temperature is approximately 5800 K.

(a)Assuming the sun's temperature is 5800 K,

The total energy of this radiation inside 1 cubic metre of sun is given by: sum V= 1m³, total energy of this radiation is given by

$U=\frac{8{\mathrm{Ï€}}^5(kT{)}^4}{15(hc{)}^3}V$

Substitute the given values

$$\begin{gathered} U=\frac{8{\mathrm{π}}^5{\left(\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)(5800\mathrm{K})\right)}^4}{15{\left(\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)\right)}^3}\left(1{\mathrm{m}}^3\right) \\ U=0.853J \end{gathered}$$

(b) Spectrum of radiation is:

$u\left(\mathrm{Ï\mu }\right)=\frac{8\mathrm{Ï€}}{(hc{)}^3}\frac{{\mathrm{Ï\mu }}^3}{e^{\mathrm{Ï\mu }/kT}-1}$

To sketch the above expression python is used and the code is:

The graph is:

The energy in terms of wavelength is:

$\mathrm{Ï\mu }=\frac{hc}{\mathrm{λ}}$

Therefore,

$$\begin{gathered} {\mathrm{Ï\mu }}_2=\frac{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}{400×{10}^{-9}\mathrm{m}}=4.9695×{10}^{-19}\mathrm{J}=3.1\mathrm{eV} \\ {\mathrm{Ï\mu }}_1=\frac{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}{700×{10}^{-9}\mathrm{m}}=2.84×{10}^{-19}\mathrm{J}=1.77\mathrm{eV} \end{gathered}$$

The code used to mark the region of the visible light

(c)Now we need to discover the component of the spectrum that represents visible light (the fraction of the energy), which we can do by integrating the following equation all across the visible light (equation 785):

$U=\frac{8\mathrm{Ï€}V}{(hc{)}^3}{∫}_{{\mathrm{Ï\mu }}_1}^{{\mathrm{Ï\mu }}_2}\frac{x^3}{e^x-1}dx$

Energy in terms of wavelength is:

$\mathrm{Ï\mu }=\frac{hc}{\mathrm{λ}}$

Hence,

$$\begin{gathered} {\mathrm{Ï\mu }}_2=\frac{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}{400×{10}^{-9}\mathrm{m}}=4.9695×{10}^{-19}\mathrm{J}=3.1\mathrm{eV} \\ {\mathrm{Ï\mu }}_1=\frac{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}{700×{10}^{-9}\mathrm{m}}=2.84×{10}^{-19}\mathrm{J}=1.77\mathrm{eV} \end{gathered}$$

Fraction of energy in visible light is:

$\frac{U_{\text{vis.}}}{U_{tot.}}=\frac{{∫}_{1.77\mathrm{eV}}^{3.1\mathrm{eV}}\frac{x^3}{e^x-1}dx}{{∫}_0^{∞}\frac{x^3}{e^x-1}dx}$

Integration in denominator equals to ${\mathrm{Ï€}}^4/15$, therefore

$\frac{U_{\text{vis.}}}{U_{tot}}=\frac{15}{{\mathrm{Ï€}}^4}{∫}_{1.77\mathrm{eV}}^{3.1\mathrm{eV}}\frac{x^3}{e^x-1}d\mathrm{Ï\mu }$

We have,

$x=\frac{\mathrm{Ï\mu }}{kT}$

The temperature is T=1500K

$$\begin{gathered} x_1=\frac{{\mathrm{Ï\mu }}_1}{kT}=\frac{1.77\mathrm{eV}}{\left(8.62×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)(5800\mathrm{K})}=3.54 \\ {x}_2=\frac{{\mathrm{Ï\mu }}_2}{kT}=\frac{3.1\mathrm{eVeV}}{\left(8.62×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)(5800\mathrm{K})}=6.8 \end{gathered}$$

Hence the integral is:

$\frac{U_{\text{vis.}}}{U_{\text{tot.}}}=\frac{15}{{\mathrm{Ï€}}^4}{∫}_{3.54}^{6.8}\frac{x^3}{e^x-1}d\mathrm{Ï\mu }$

To evaluate the integral, python is used, the code is given below and the ratio is:

$\frac{U_{\text{vis.}}}{U_{tot.}}=0.382$