Asked by Vinie
What is the energy of a photo released when a Li^2+ is relaxed from 5 state to 4 state
Answers
Answered by
DrBob222
Do you mean from n = 5 to n = 4?
1/wavelength = RZ^2[1/(n1)^2 - 1/(n2)^2]
R = 1.0973732E^7/meter
Z = 3 for Li
n1 = 4
n2 = 5
Solve for wavelength, then
E = hc/wavelength where E is the energy in joules, h = Planck's constant = 6.6262E-34 Js
c = speed of light 3E8 m/s
Post your work if you get stuck.
Yes, there IS an equation that does all of this in one step but I don't remember it. OR, from what I've written you can derive it.
1/wavelength = RZ^2[1/(n1)^2 - 1/(n2)^2]
R = 1.0973732E^7/meter
Z = 3 for Li
n1 = 4
n2 = 5
Solve for wavelength, then
E = hc/wavelength where E is the energy in joules, h = Planck's constant = 6.6262E-34 Js
c = speed of light 3E8 m/s
Post your work if you get stuck.
Yes, there IS an equation that does all of this in one step but I don't remember it. OR, from what I've written you can derive it.
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