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In the lecture today, when we were given the example of calculating the frequency of light emitted we used the change of E=Efinal- EInitial and we got -4.09x10^-19j then when we used E=hv and changed the answer we got to a positive 4.09x10^-19j. Why is that? Thank you.
The whole idea of conservation of energy is that energy can't be lost or destroyed. So when the e- is giving off energy, it has to go somewhere. In this case the energy lost is being emitted as light. It's positive because it's the same exact amount of energy, but it's not being lost here- it's being emitted. Hope this makes sense!
Hey! I posted a reply to this in Lec 7. Question [ENDORESED] but I've attached it below. I hope it helps!
Lucy Wang 1A wrote:The negative sign simply is to show that this is the amount of energy LOST by the electron. We changed it to a positive sign to show the amount of energy being ABSORBED by the photon or turned into light. This works because according to the conservation of energy, energy cannot be created or destroyed. So basically, while the electron is losing this energy, it is passing it on to the photon which allows the energy lost from the electron to be shown as light.
To give an analogy, its like making a payment to someone. In your account (electron) it may show -$2.00, but in their account (photon) it would be +$2.00. lol not sure if that makes sense but I hope it was helpful
Hi! In the equation En=(-hR)/(n)^2, the negative sign refers to the decrease in the electron's energy after it returns to the ground state and emits electromagnetic radiation, thus losing energy. So if the question had asked how much energy the electron lost, the answer would be the -4.09x10^-19 J. However, because the question asked what the frequency was of the light emitted by the electron, we use the positive form of the difference in energy to denote that the energy of the light released is positive. Essentially, it shows the transfer of energy and that the energy is positively transferred as the light emitted but negatively taken away from the electron that emitted the light. Hope this helps!
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