#42: An excited hydrogen atom emits light with a frequency of 1.14 x 10^14 Hz to reach the energy level n = 4. In what principle quantum level did the electron begin?
A. n = 5
B. n = 6
C. n = 4
D. n = 7
How do you do this problem (step by step)??
Post Assessment Module #42
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Sean Monji 2B
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Re: Post Assessment Module #42
1. Through the emitted photon, we can figure out the net change of energy of the electron.
E = hv
2. Using this energy, we have all the variables for the E = -RH/n^2[initial] - (-RH/n^2) [final] but must solve for n [initial]
We know that the energy level must have gone from a higher number to 4 because energy was emitted, thus the e- lost energy and went to a lower energy state (also the answers indicate this)
3. Derived equation -> n [initial] = 1 / (-E/Rh + 1/n^2)^ .5
4. Plug in and solve: 1 / (-1.14 * 10 ^14 / R + 1 / 16 ) ^ .5
(you can substitute hv for E while factoring out -Rh, to make hv = -Rh(1 / n^2 - 1 / n^2) and get n = 1 / (-v / R + 1 / n^2) ^ .5 so that the h cancels out)
I got n[initial] = 6
Hope that makes sense
E = hv
2. Using this energy, we have all the variables for the E = -RH/n^2[initial] - (-RH/n^2) [final] but must solve for n [initial]
We know that the energy level must have gone from a higher number to 4 because energy was emitted, thus the e- lost energy and went to a lower energy state (also the answers indicate this)
3. Derived equation -> n [initial] = 1 / (-E/Rh + 1/n^2)^ .5
4. Plug in and solve: 1 / (-1.14 * 10 ^14 / R + 1 / 16 ) ^ .5
(you can substitute hv for E while factoring out -Rh, to make hv = -Rh(1 / n^2 - 1 / n^2) and get n = 1 / (-v / R + 1 / n^2) ^ .5 so that the h cancels out)
I got n[initial] = 6
Hope that makes sense
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