Question
Rank these photons in terms of decreasing energy: IR ( $\nu=$ $\left.6.5 \times 10^{13} \mathrm{~s}^{-1}\right) ;$ microwave $\left(\nu=9.8 \times 10^{11} \mathrm{~s}^{-1}\right) ; \mathrm{UV}\left(\nu=8.0 \times 10^{15} \mathrm{~s}^{-1}\right)$
Step 1
According to Planck's equation, the energy of a photon is directly proportional to its frequency. This means that a photon with a higher frequency will have a higher energy. Show more…
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Rank these photons in terms of decreasing energy: (a) IR $\left(v=6.5 \times 10^{13} \mathrm{s}^{-1}\right) ;$ (b) microwave $\left(v=9.8 \times 10^{11} \mathrm{s}^{-1}\right);$ (c) $\mathrm{UV}\left(v=8.0 \times 10^{15} \mathrm{s}^{-1}\right)$
Show that the photons in a 1240 -nm infrared beam have energies of $1.00 \mathrm{eV}$ $$ \mathrm{E}=h f=\frac{h \mathrm{c}}{\lambda}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{1240 \times 10^{-9} \mathrm{~m}}=1.602 \times 10^{-19} \mathrm{~J}=1.00 \mathrm{eV} $$
What is the energy of a photon corresponding to microwave radiation of frequency $1.258 \times 10^{10} / \mathrm{s} ?$
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