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.
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