# Basic Quantum mechanics, H2 approximation with SHO

## 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

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vela
Staff Emeritus
Homework Helper
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

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

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.

vela
Staff Emeritus