Answer
The energy of photons emitted by the AM radio station is $6.692\times10^{-28}J$.
The energy of photons emitted by the FM radio station is $6.513\times10^{-26}J$.
Therefore, the energy of photons emitted by the FM radio station is higher than that by the AM radio station.
Work Step by Step
Strategy:
The following formula would be used to answer this question. $$E=h\times\nu$$
in which
$E$ : the energy of a photon emitted by a radio station
$h$ : Planck constant ($h\approx6.626\times10^{-34}J.s$)
$\nu$ : frequency of the photons emitted
We would calculate the energy of photons in each case and compare them after that.
- Step 1: Find the known variables
The AM radio station broadcasts at $1010kHz$; so the frequency of photons emitted $\nu_{AM}=1010kHz=1.01\times10^6s^{-1}$.
The FM radio station broadcasts at $98.3MHz$; the frequency of photons emitted $\nu_{FM}=98.3MHz=9.83\times10^7s^{-1}$.
(For those who may not know, $1MHz = 10^3kHz=10^6Hz=10^6s^{-1}$)
- Step 2: Calculate the energy of photons emitted by radio stations
The AM radio station:
$E_{AM}=h\times\nu_{AM}=(6.626\times10^{-34})\times(1.01\times10^6)\approx6.692\times10^{-28}J$
The FM radio station:
$E_{FM}=h\times\nu_{FM}=(6.626\times10^{-34})\times(9.83\times10^6)\approx6.513\times10^{-26}J$
The energy of photons emitted by the FM radio station is higher than that by the AM radio station.
