# Basic Quantum mechanics, H2 approximation with SHO

• osheari1
In summary: This takes into account the masses of both particles and their relative distance from the center of mass. Once you have the correct reduced mass, you can use it to calculate the energy levels and wavelengths accurately.

## 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 can't 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

osheari1 said:
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 can't 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

The first excited state is n=2, so the third excited state is n=?

ahh right, H doesn't 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.

You need to calculate the reduced mass, which is the effective mass of the oscillator.