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Help with Quantum Mechanics.
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Hey, If anyone can help me with this it'd be much appreciated, tnx.
*Q*:A hydrogen atom is in the third excited state. It makes a transition to a different state and a photon is either absorbed or emitted. Determine the quantum number nf of the final state and the energy of the photon when the photon is (a) emitted with the shortest possible wavelength (b) emitted with the longest possible wavelength and (c) absorbed with the longest possible wavelength.
*Q*:A hydrogen atom is in the third excited state. It makes a transition to a different state and a photon is either absorbed or emitted. Determine the quantum number nf of the final state and the energy of the photon when the photon is (a) emitted with the shortest possible wavelength (b) emitted with the longest possible wavelength and (c) absorbed with the longest possible wavelength.
Answers
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Here are a few clues:
The base state energy is E_1=-m x e^4/(2 x hbar^2)
x here means multiply and hbar is Planck's constant/2 x pi
The energy of higher excited states, say the third, is equal to E_
The base state energy is E_1=-m x e^4/(2 x hbar^2)
x here means multiply and hbar is Planck's constant/2 x pi
The energy of higher excited states, say the third, is equal to E_
08:57 Sat 26th Feb 2011
Here are a few clues:
The base state energy is E_1=-m x e^4/(2 x hbar^2)
x here means multiply and hbar is Planck's constant/2 x pi
The energy of higher excited states, say the third, is equal to E_1/4^2=E_1/16
Generally the nth excited state has an energy equal to E_1/(n+1)^2
so this means that the energies get larger quite quickly as n increases because don't forget that E_1 is negative.
The quantum number is n above.
When light is absorbed the electron moves to a higher orbit and when light is emitted it moves to a lower orbit.t
The frequency is related to the change in energy by
h x frequency=change in energy.
The base state energy is E_1=-m x e^4/(2 x hbar^2)
x here means multiply and hbar is Planck's constant/2 x pi
The energy of higher excited states, say the third, is equal to E_1/4^2=E_1/16
Generally the nth excited state has an energy equal to E_1/(n+1)^2
so this means that the energies get larger quite quickly as n increases because don't forget that E_1 is negative.
The quantum number is n above.
When light is absorbed the electron moves to a higher orbit and when light is emitted it moves to a lower orbit.t
The frequency is related to the change in energy by
h x frequency=change in energy.
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