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Question

Estimate the number of photons emitted per second from 1.0cm21.0 \mathrm{cm}^{2} of a person’s skin if a typical emitted photon has a wavelength of 10,000 nm.

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The area of the sample skin is given as

A=1 cm2=1×10−4 m2\begin{align*} A & = 1 \ \mathrm{cm^2 } = 1 \times 10^{-4} \ \mathrm{m^2 } \end{align*}

The wavelength of emitted photon is given as

λmax=10,000 nm=1×10−5 m\begin{align*} \lambda_{max} & = 10,000 \ \mathrm{nm} = 1 \times 10^{-5} \ \mathrm{m } \end{align*}

Now, the temperature of the human body, using Wien's law, is estimated as

λmax=2.90×10−3 m⋅KT  ⟹  T=2.90×10−3 m⋅Kλmax\begin{align*} \lambda_{max} & = \dfrac{2.90 \times 10^{-3} \ \mathrm{m\cdot K}}{T} \\ \implies T & = \dfrac{2.90 \times 10^{-3} \ \mathrm{m\cdot K}}{\lambda_{max} } %\\ %& = \dfrac{2.90 \times 10^{-3} \ \mathrm{m\cdot K}}{(1.0 \times 10^{-5} \ \mathrm{m})} \\ %& \approx 290 \ \mathrm{K} \end{align*}

Stefan-Boltzmann's constant is given as

σ=5.67×10−8 W/m2⋅K4\begin{align*} \sigma & = 5.67 \times 10^{-8} \ \mathrm{W/m^2\cdot K^4} \end{align*}

If nn number of photons are emitted per second, then the power is given as

P=n⋅hcλmax\begin{align*} P & = n \cdot \dfrac{h c}{\lambda_{max}} \end{align*}

Using Stefan-Boltzmann's law, we have the emitted power as

P=σ A T4\begin{align*} P & = \sigma \, A\, T^4 \end{align*}

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