Answer to Question #108339 in Quantum Mechanics for Shivi

Question

Consider a proton as a bound oscillator with a natural frequency of
4 ×10 Hz. 21 
Calculate the energy of its ground and first excited states.

Answer 1

The energy levels of bound harmonic oscillator are

$(1)E_n=h\nu\cdot (n+\frac{1}{2}), n=0,1,2,...$

The ground state must correspond to a stable particle - the proton and its rest energy. Proton has rest energy $ϵ=938MeV$ This value corresponds to the oscillator frequency, which can be found from the equality of the ground state energy and the photon quantum

$\frac12​hν=ϵ;$

$ν=\frac{2ϵ}h​=\frac{9,38\times10^8eV}{4.14\times10^{−15}eVs}​=4.53\times10^{23}Hz$

The value of the Planck constant bar expressed in various quantities can be found in [1].

The first excited states have energy $E_n=\frac{3}{2}\cdot h\nu=3\epsilon=2.81\cdot 10^3 MeV$

Answer: The energy of ground and first excited states of proton are 938 MeV

and $2.81\times10^3MeV$