jdstokes
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Hi all,
I have an exam tomorrow and this problem concerns me greatly.
An electron is located in an infinitely deep one-dimensional square potential well. The width of the well is 1.00 nm.
(e) Light is shone on the electron causing it to jump from the ground state to the n = 3 state. What is the wavelength of the light?
(f) What would happen if light with wavelength twice that calculated in part f were shone on the electron in its ground state.
Answer to Question (f):
This does not correspond to a transition energy from the ground state to a higher energy state and so the photons will not be absorbed.
I'll work through the problem algebraically to show why I think the photon will in fact be absorbed.
The energy of the nth state is E_n = n^2E_1 where E_1 = \frac{h^2}{8mL^2} is the energy of the ground state.
The transition energy between n = 3 and and the ground state is thus
\Delta E = E_3 - E_1 = 3^2E_1 - E_1 = 8E_1 = \frac{h^2}{mL^2}.
So the wavelength of the incident photon is
\lambda = \frac{hc}{\Delta E}.
If the light in part (f) has twice this wavelength, then the energy of the photons is half the value in part (e), ie
\Delta E' = 1/2\Delta E = 4E_1,
Since the transition energy between the ground state and the n = 2 state is just 3E_1, why won't the electron be promoted to the n = 2 state?
Thanks.
James
I have an exam tomorrow and this problem concerns me greatly.
An electron is located in an infinitely deep one-dimensional square potential well. The width of the well is 1.00 nm.
(e) Light is shone on the electron causing it to jump from the ground state to the n = 3 state. What is the wavelength of the light?
(f) What would happen if light with wavelength twice that calculated in part f were shone on the electron in its ground state.
Answer to Question (f):
This does not correspond to a transition energy from the ground state to a higher energy state and so the photons will not be absorbed.
I'll work through the problem algebraically to show why I think the photon will in fact be absorbed.
The energy of the nth state is E_n = n^2E_1 where E_1 = \frac{h^2}{8mL^2} is the energy of the ground state.
The transition energy between n = 3 and and the ground state is thus
\Delta E = E_3 - E_1 = 3^2E_1 - E_1 = 8E_1 = \frac{h^2}{mL^2}.
So the wavelength of the incident photon is
\lambda = \frac{hc}{\Delta E}.
If the light in part (f) has twice this wavelength, then the energy of the photons is half the value in part (e), ie
\Delta E' = 1/2\Delta E = 4E_1,
Since the transition energy between the ground state and the n = 2 state is just 3E_1, why won't the electron be promoted to the n = 2 state?
Thanks.
James