Answer to Question #143399 in Quantum Mechanics for aks

Question

6. A sodium atom is in one of the states labeled ''Lowest excited levels". It remains in that state
for an average time of 1.610-8
s before it makes a transition back to a ground state, emitting
a photon with wavelength 589.0 nm and energy 2.105 eV. What is the uncertainty in energy
of that excited state? What is the wavelength spread of the corresponding spectrum line?

Answer 1

$\Delta E\Delta t\geqslant\frac{\hbar }{2}$

$\Delta E\geqslant\frac{\hbar }{2\Delta t}=\frac{1.05\cdot10^{-34}}{2\cdot1.6\cdot10^{-8}}=3.3\cdot10^{-27}$ J

$\Delta \lambda \geqslant \frac{\hbar }{2\Delta p}=\frac{\hbar c}{2\Delta E}=\frac{\hbar c}{2(\frac{hc}{\lambda }-E_0)}=\frac{\hbar c\lambda}{2({hc}-E_0\lambda)}=\frac{1.054\cdot10^{-34}\cdot3\cdot10^{8}\cdot589\cdot10^{-9}}{2\cdot(6.63\cdot10^{-34}\cdot3\cdot10^{8}-2.105\cdot1.6\cdot10^{-19}\cdot589\cdot10^{-9})}=17.7\cdot10^{-6}$ m