Quantum Chemistry/Example 34

Calculate the wavenumber corresponding to the transition of a hydrogen atom from the n=2 ℓ =1 m_ℓ =-1 state to the n=2 ℓ =1 m_ℓ =1 state in a 1 T magnetic field.

Answer

First, the equation for energy of a hydrogen atom under a magnetic field is needed. The energy of a hydrogen atom under a magnetic field can be expressed as follows:

$E=-{\frac {m_{e}e^{4}}{8\varepsilon _{0}h^{2}n^{2}}}+\beta _{B}m_{l}B_{z}$

Where E is the energy in joules, m_e is the mass of the electron in kg, e is the charge of electron in Coulomb “C”, ε₀ is vacuum permittivity in F/m, h is Planck’s constant in J·s, and n is the principal quantum number. In the other half of the equation, which considers the magnetic quantum number, B_z is the magnetic field in Tesla “T”, m_l is the magnetic quantum number, and β_B is Bohr magneton in J/T, and it is calculated using the following equation:

$\beta _{B}={\frac {|e|\cdot \hbar }{2m_{e}}}={\frac {|1.6022\times 10^{-19}C|\cdot 1.0546\times 10^{-34}J\cdot s}{2(9.1094\times 10^{-31}kg)}}=9.2740\times 10^{-24}J/T$

Therefore, using the equation mentioned for the energy of a hydrogen atom, the energy of the transition, which is the difference in energy “∆E”, can be described as follows:

$\Delta E=E_{upper}-E_{lower}=(-{\frac {m_{e}e^{4}}{8\varepsilon _{0}h^{2}n_{upper}^{2}}}+\beta _{B}m_{l,upper}B_{z})-(-{\frac {m_{e}e^{4}}{8\varepsilon _{0}h^{2}n_{lower}^{2}}}+\beta _{B}m_{l,lower}B_{z})$

If we go back to the question we can notice that there is no change in the quantum number “n” as “n=2” in both states. Also, if we look at the first term of the energy equation, we can see that it all consists of constants except the quantum number “n”. Thus, since “n” is the same for both states, both terms can get cancelled. Therefore, the equation becomes:

$\Delta E=\beta _{B}m_{l,upper}B_{z}-\beta _{B}m_{l,lower}B_{z}$

Thus, after substituting:

$\Delta E=\beta _{B}m_{l,upper}B_{z}-\beta _{B}m_{l,lower}B_{z}=\beta _{B}B_{z}(m_{l,upper}-m_{l,lower})=9.27\times 10^{-24}J\cdot T^{-1}(1T)[1-(-1)]=9.27\times 10^{-24}J\cdot T^{-1}(1T)(2)=1.85\times 10^{-23}J$

Then, we can convert the energy to wavenumber:

$E=hc{\tilde {\nu }}\Longrightarrow {\tilde {\nu }}={\frac {E}{hc}}={\frac {1.85\times 10^{-23}J}{6.626\times 10^{-34}Js(3.00\times 10^{8}m/s)}}=93.3m^{-1}=0.933cm^{-1}$

This value is reasonable as the difference between energies/wavenumbers of transitions is usually small. The instrument that was used for the transition in the question is electron paramagnetic resonance (EPR) spectroscopy, which is a famous type of spectroscopy that measure the absorption of the EM radiation, but it does it using magnetic filed, and it does apply that magnetic field on the unpaired electrons. EPR spectroscopy specifically measures the absorption of microwave radiation, which is approximately in the the range of ~0.1-10.0cm^-1. Thus, the answer does make sense.