Example during lecture 6
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Example during lecture 6
In the example problem prof went over during Friday's (10/16) lecture, he removed the negative sign from the -4.01 x 10^-19 J answer when plugging it in to the next equation. Could someone explain why he did that? I didn't quite understand
Re: Example during lecture 6
I wish I could help you with that- I didn't realize he removed the negative sign. Maybe the next equation had a negative in front of that variable and they cancelled out?
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Re: Example during lecture 6
Hi,
I believe in the example problem, we are considering the hydrogen atom as the system. Because the 4.01 x 10^-19 joules is released from the atom, it is seen as a loss of energy in the system and is thus labeled negative. However the second question asks to find the frequency of the light emitted, making the light released the thing we are focusing on. Light is energy, so in this situation 4.01 x 10^-19 J would have to be positive because the light cannot have negative energy.
Hope this helps
I believe in the example problem, we are considering the hydrogen atom as the system. Because the 4.01 x 10^-19 joules is released from the atom, it is seen as a loss of energy in the system and is thus labeled negative. However the second question asks to find the frequency of the light emitted, making the light released the thing we are focusing on. Light is energy, so in this situation 4.01 x 10^-19 J would have to be positive because the light cannot have negative energy.
Hope this helps
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Re: Example during lecture 6
Shannon Moore 2L wrote:In the example problem prof went over during Friday's (10/16) lecture, he removed the negative sign from the -4.01 x 10^-19 J answer when plugging it in to the next equation. Could someone explain why he did that? I didn't quite understand
Dylan gave a good answer, but I wanted to expand a little more on what was going on.
For the spectrum given for hydrogen (and for any spectrum), a decrease in n results in a decrease in energy in the electron. Think of it as a ball at the top of the hill. At the top of the hill, the ball has much energy, and wants to roll down the hill. So, even though the ball moves, its energy decreases at the end of the process. For the example given, think of n = 4 as higher up on the hill and think of n = 2 as lower on the hill. The total potential energy of the ball at n = 4 is higher than the potential energy of the ball at n = 2, so when going from n = 4 to n = 2, there was a net decrease in energy, which is why there was a negative sign in front of -4.01 x 10^-19 J.
Once again looking at the problem in the context of the ball rolling down the hill, we must recognize that the energy in that ball is changing. When going down the hill, the ball is converting gravitational potential energy to kinetic energy. So, the gravitational potential energy is decreasing (the negative sign in front of -4.01 x 10^-19) while the kinetic energy of the ball is increasing (positive 4.01 x 10^-19). The second part of the example he gave is analogous to the kinetic energy of the ball. He used the energy he calculated from going from n = 4 to n = 2 as a way to find out how much energy was emitted by the electron, much like how one can calculate the kinetic energy of the ball from the ball's change in gravitational potential energy. So, think of the first part as the gravitational energy of the ball (decreasing, so negative) and the second part of the question as the kinetic energy (equal, because of conservation of energy, but positive).
Of course, it is also important to remember that the second part, the positive part, was in reference to the emitted light, and didn't directly have to do with the electron itself (unlike how the ball's energy is converted, but stays as a part of the ball system). However, the analogy stays mostly the same.
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Re: Example during lecture 6
If an electron jumps from a high state of energy to a low energy state, the change in energy is negative because this process release energy. However, when we use the energy later to calculate either the frequency or the wavelength, we usually do not include the negative sign, or you can understand as we usually use the absolute value. This is because wavelength and frequency are always reported as positive quantities. This is what word "convention" means in that lecture.
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Sapling Week 2 Question 9
What is the easiest way to calculate wavelength when an atom is transition between different energy levels (levels for n)?
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