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Basic Quantum mechanics, H2 approximation with SHO

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Homework Statement



A H2 molecule can be approximated by a simple harmonic oscillator having spring constant k = 1.1*10^3 N/m. Find a() the energy levels, and (b) the possible wavelengths of photons emitted when the H2 molecule decays from the third excited state eventually to the ground state.

Homework Equations



En = ( n + 1/2 ) h_bar*ω

w^2 = k/m

The Attempt at a Solution



I solved for omega by √(1.1E3/(2*(mass of electron(kg) + mass of proton + mass of neutron))
then multiplied by the eV version of h_bar and got En=(n+1/2).2668 eV

However the book says its En=(n+1/2).755eV


I tried using the books answer to solve for the mass, and got 8.53E-28 kg but I cant see where they would be getting that answer.





However, I tried solving party b assuming the books answer was correct


First I solved for each energy level drop

E_3→1 = (3+1/2).755 - (1+1/2).755 = 1.52 eV corresponding λ = 815.8 nm books answer = 549 nm

E_3→2 = (3+1/2).755 - (2+1/2).755 = .76 eV corresponding λ = 1631.6 nm books λ=821 nm

E_2→1 = (2+1/2).755 - (1+1/2).755 = .76 eV corresponding λ = 1631.58 nm books λ = 1640 nm



please help me solve this problem Im quite confused
 

Answers and Replies

  • #2
vela
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I solved for omega by √(1.1E3/(2*(mass of electron(kg) + mass of proton + mass of neutron))
then multiplied by the eV version of h_bar and got En=(n+1/2).2668 eV
Why are you using the mass of a neutron? H2 doesn't have any neutrons.
However the book says its En=(n+1/2).755eV


I tried using the books answer to solve for the mass, and got 8.53E-28 kg but I cant see where they would be getting that answer.
You need to use the reduced mass. Do you know how to calculate that?

However, I tried solving party b assuming the books answer was correct


First I solved for each energy level drop

E_3→1 = (3+1/2).755 - (1+1/2).755 = 1.52 eV corresponding λ = 815.8 nm books answer = 549 nm

E_3→2 = (3+1/2).755 - (2+1/2).755 = .76 eV corresponding λ = 1631.6 nm books λ=821 nm

E_2→1 = (2+1/2).755 - (1+1/2).755 = .76 eV corresponding λ = 1631.58 nm books λ = 1640 nm



please help me solve this problem Im quite confused
The first excited state is n=2, so the third excited state is n=?
 
  • #3
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ahh right, H doesnt have neutrons
However, even when I only use protons and electrons I get a wrong answer







and I realize my mistake for the excited states now.
 
  • #4
vela
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You need to calculate the reduced mass, which is the effective mass of the oscillator.
 

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