# Quantum Chemistry/Example 10

## Question

Use Planck's radiation law to find the surface temperature of the Sun when its maximum intensity of EM radiation is emitted at 504 nm. Ensure the temperature units are in Kelvins.

### Getting the correct function

A blackbody material is defined by its ability to absorb ALL radiation that falls onto it. When the blackbody is at constant temperature, the distribution of its emission frequency can be determined by assuming its only direct relation is to temperature. Furthermore, the frequency of the electromagnetic (EM) radiation can also be measured in units of wavenumber ${\text{cm}}^{-1}$ .

{\begin{aligned}\rho (\nu ,T)&={\frac {2h\nu ^{2}}{c^{2}}}\left\langle E(\nu ,T)\right\rangle &&\rho (\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}\left\langle E(\nu ,T)\right\rangle \end{aligned}}

Planck's approach to defining ${\textstyle \langle E(v,T)\rangle }$  was deriving for a closed form Harmonic Oscillator was based on Boltzmann's distribution. The resulting form of Planck's law can then be applied to the question now that all parameters are known except for the parameter in question. The model below shows the radiation intensity distribution (i.e., area of the curve) of EM with respect to frequency (in Hertz; Hz) and temperature (in Kelvins; K).

{\begin{aligned}d\rho (v,T)&=\rho (\nu ,T)d\nu {\text{, where }}\rho (\nu ,T)=\left({\frac {2h\nu ^{3}}{c^{2}}}\right)\left[exp\left({\frac {h\nu }{k_{B}T}}\right)\right]^{-1}\\&{\text{Note that }}\left|{\frac {d\nu }{d\lambda }}\right|={\frac {c}{\lambda ^{2}}}\\d\rho (\lambda ,T)&=\rho (\lambda ,T)d\lambda {\text{, where }}\rho (\lambda ,T)=\left({\frac {2hc^{2}}{\lambda ^{5}}}\right)\left[exp\left({\frac {hc}{k_{B}T\lambda }}\right)\right]^{-1}\end{aligned}}

Planck's law for a blackbody material predicts the behaviour of its quantitative properties (such as frequency, wavelength, and temperature) when the environment are in extreme conditions.
 Planck's Law of BlackBody Radiation $\rho (\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}\times {\frac {1}{\left[exp({\frac {hc}{k_{B}\lambda T}})\right]-1}}$ ### Solving for the temperature

The maximum value of the independent variable of any function can be found by equating the function's derivative to zero. Only $T$  is unknown and is with respect to the independent variable, $\lambda$ . Thus, Planck's law will be derived with respect to $\lambda$ .

First, for simplicity, let ${\textstyle x={\frac {hc}{k_{B}\lambda T}}}$ .

${d\rho \over d\lambda }=(2hc^{2})\left[{\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}-{\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}\right]=0$

The first term consists only of non-zero constants so that leaves:

${\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}-{\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}=0$

{\begin{aligned}{\frac {x}{\lambda ^{6}}}{\frac {e^{x}}{\left(e^{x}-1\right)^{2}}}&={\frac {5}{\lambda ^{6}\left(e^{x}-1\right)}}\\{\frac {xe^{x}}{e^{x}-1}}&=5\\{\frac {xe^{x}}{e^{x}-1}}-5&=0\\(x-5)e^{x}+5&=0\end{aligned}}

Subbing ${\textstyle x={\frac {hc}{k_{B}\lambda T}}}$  back in gives:

{\begin{aligned}\left({\frac {hc}{k_{B}\lambda T}}-5\right)e^{x}&=0&&\rightarrow &&&{\frac {hc}{k_{B}\lambda T}}-5=0\end{aligned}}

At max intensity, ${\textstyle \lambda _{max}=504{\text{nm}}}$ . Subbing in all known values and constants (in SI units) then rearranging to solve for T will determine the surface temperature of the Sun under these specific conditions.

Known Values
Variable Value Units
$h$  $6.62607\times 10^{-34}$  $m^{2}kg \over s$
$\lambda _{max}$  $504\times 10^{-9}$  $m$
$k_{B}$  $1.38065\times 10^{-23}$  ${\frac {m^{2}kg}{s^{2}K}}$
$c$  $2.99792\times 10^{8}$  $m \over s$

{\begin{aligned}T={\frac {hc}{5k_{B}\lambda _{max}}}={\frac {(6.62607\times 10^{-34})(2.99792\times 10^{8})}{5(1.38065\times 10^{-23})(504\times 10^{-9})}}=5709.432{\text{ }}K\end{aligned}}

Therefore, the surface temperature of the Sun at a maximum wavelength of 504 nm is just above 5709 Kelvins.