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Ch.7 - Quantum-Mechanical Model of the Atom

Chapter 7, Problem 42

Calculate the energy of a photon of electromagnetic radiation at each of the frequencies indicated in Problem 40. a. 100.2 MHz (typical frequency for FM radio broadcasting) b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures) c. 835.6 MHz (common frequency used for cell phone communication)

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hey everyone in this example, we need to find the corresponding energy of a photon for the below frequencies. So we should recall our formula for the energy of a photon, which is equal to Planck's constant. Multiplied by our frequency of our photon. We should recall that frequency is going to be in units of hertz or is equivalent to also inverse seconds. So beginning with part A, we have that energy is equal to Planck's constant. We should recall is 6.626 times 10 to the negative 34th power in units of jewels times seconds. Then this is multiplied by our frequency, which we should recognize is given in gigahertz. And because our prefixes giga, that tells us that we have 10 to the ninth. Power inverse seconds or hertz. And so we would say that we have a frequency of 2.2 times 10 to the ninth Power inverse seconds. Because we want to go ahead and cancel out our units of seconds. And this leaves us with jewels as our final unit for energy. So what we're going to get when we multiply that out Is a value of 1. times 10 to the negative 24th power jewels for our first given frequency in part A. And so this is our first answer. Moving on to Part B, We have that energy is again equal to planck's constant, which we recall is 6.626 times 10 to the negative 34th power jewels times seconds. And then we want to multiply by our frequency and units of inverse seconds. So our prefix tera in part B tells us that we have 10 to the 12th power inverse seconds As our frequency. So that would be plugged in as 4.2 times 10 to the 12th power inverse seconds, allowing us to cancel out our units of seconds. Leaving us with jewels yet again for energy. And what we're going to get is a value of 2.8 times 10 to the negative 21st power jewels. And so this would be our second answer um as energy for the given frequency and part B. Moving on to part C we have our energy of our photon is equal to Planck's constant for the last time 6.626 times 10 to the negative 34th power jewels time seconds. And then multiplied by are given frequency in units of megahertz are prefix mega is telling us that we have 10 to the six power inverse seconds. And so we would plug that in as a frequency of 102. Times 10 to the six power inverse seconds, allowing us to cancel our units of seconds. Leaving us with jewels for our energy of our photon. And we're going to get a value equal to 2.8 times 10. Sorry about that. We would get a value of 6.791 times to the negative 26 power jewels. And so this would be our final answer for our third example, corresponding to its photons energy based on the given frequency in megahertz. So everything boxed in blue are are three final answers. If you have any questions, please leave them down below, and I will see everyone in the next practice video.