## Chemistry: The Central Science (13th Edition)

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.
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.