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An AM radio station broadcasts at 1010 kHz, and its FM partner broadcasts at 98.3 MHz. Calculate and compare the energy of the photons emitted by these two radio stations.
Solution
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The wavelength of AM radio station is $v_{AM} = 1010 \mathrm{kHz} = 1010 \cdot 10^{3} \mathrm{Hz} = 1010 \cdot 10^{3} \mathrm{s^{1}}$

The wavelength of FM radio station is $v_{FM} = 98.3 \mathrm{MHz} = 98.3 \cdot 10^{10^{6}} \mathrm{s^{1}}$
The Plank's constant is $h = 6.626 \cdot 10^{34} \mathrm{J \cdot s}$
Let us find the energy of photons emitted by these two radio stations.
$\bullet$ The energy of AM radio stations is
$\begin{align*} E_{AM} &= h \cdot v_{AM}\\ &= 6.626 \cdot 10^{34} \mathrm{J \cdot s} \cdot 1010 \cdot 10^{3} \mathrm{s^{1}}\\ &= {\color{#4257b2}6.69 \cdot 10^{28} \mathrm{J}} \end{align*}$
$\bullet$ The energy of FM radio stations is
$\begin{align*} E_{FM} &= h \cdot v_{FM}\\ &= 6.626 \cdot 10^{34} \mathrm{J \cdot s} \cdot 98.3 \cdot 10^{10^{6}} \mathrm{s^{1}}\\ &= {\color{#4257b2}6.51 \cdot 10^{26} \mathrm{J}} \end{align*}$