The energy of electromagnetic waves can be calculated by using the following formula:
\(E=\frac{hc}{\lambda}\)
where:
\(h=6.63 \cdot 10^{-34} Js\) is the Planck constant
\(c=3\cdot 10^8 m/s\) is the speed of light
\(\lambda\) is the wavelength of the wave
Substituting the various wavelengths in the formula, we find:
IR-C:
\(\lambda=3000 nm=3000\cdot 10^{-9} m\) -->
\(E=\frac{hc}{\lambda}=\frac{(6.63 \cdot 10^{-34})(3 \cdot 10^8)}{3000\cdot 10^{-9}}=6.63 \cdot 10^{-20} J\)
\(\lambda=1,000,000 nm=1\cdot 10^{-3} m\) -->
\(E=\frac{hc}{\lambda}=\frac{(6.63 \cdot 10^{-34})(3 \cdot 10^8)}{1\cdot 10^{-3}}=1.99 \cdot 10^{-22} J\)
IR-A:
\(\lambda=700 nm=700\cdot 10^{-9} m\) -->
\(E=\frac{hc}{\lambda}=\frac{(6.63 \cdot 10^{-34})(3 \cdot 10^8)}{700\cdot 10^{-9}}=2.84 \cdot 10^{-19} J\)
\(\lambda=1400 nm=1400\cdot 10^{-9} m\) -->
\(E=\frac{hc}{\lambda}=\frac{(6.63 \cdot 10^{-34})(3 \cdot 10^8)}{1400\cdot 10^{-9}}=1.42 \cdot 10^{-19} J\)
Therefore, we see that IR-A have higher energy than IR-C.
