Hi all.
What is the thought process for understanding the second part of the question "Considering that microwave ovens radiate at 2.45 GHz, how would you respond to the debate"? I guessed on it since I wasn't sure what I was looking for.
Achieve HW #7
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Re: Achieve HW #7
From what I understood, you convert 2.45GHz to meters. First, you would convert 2.45 GHz to Hz by multiplying by 10^9 (giga=10^9) to get 2.45 x 10^9 Hz. Then you convert the frequency to meters using λ=c/v .
If the wavelength of the microwave is larger than the wavelength you calculated for the first part of the problem, then the microwave cannot eject electrons from the metal because it will not have enough energy to overcome the work function of the metal (the energy required to remove the electron).
If the wavelength of the microwave is larger than the wavelength you calculated for the first part of the problem, then the microwave cannot eject electrons from the metal because it will not have enough energy to overcome the work function of the metal (the energy required to remove the electron).
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Re: Achieve HW #7
Hi,
So the idea behind this question is to understand that in order for an electron to be ejected, the radiation must reach a minimum frequency, and this is determined based on each specific metal. So considering that microwave ovens radiate at 2.45 GHz, and the work function given above in the question. You can determine whether or not the 2.45GHz is above the frequency threshold of the metal, as per the photoelectric effect.
So the idea behind this question is to understand that in order for an electron to be ejected, the radiation must reach a minimum frequency, and this is determined based on each specific metal. So considering that microwave ovens radiate at 2.45 GHz, and the work function given above in the question. You can determine whether or not the 2.45GHz is above the frequency threshold of the metal, as per the photoelectric effect.
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