LOG#169. A two level problem.

lasersword Laser_beam

This post is the solution of the following problem: pick an atom with two energy levels. They have a transition wavelength of 580nm. At room temperature 5\cdot 10^{20}atoms are in the lower state.

1) How many atoms are in the upper state, if we have thermal equilibrium?

2) Suppose instead we had 9\cdot 10^{20} atoms pumped into the upper state, with 5\cdot 10^{20} atoms in the lower state. This is a non-equilibrium state. How much energy (in Joules) could be released in a single pulse of light if we restored the equilibrium?

Data: Boltzmann factor is \exp\left(-E/k_BT \right) and k_B=1\mbox{.}38\cdot 10^{-23}J/K

Solution to 1). If N_b is the higher energy state and N_a is the lower energy state, then the energy difference between these two atomic states can be calculated using the transition wavelength (580nm), i.e.,

    \[E_b-E_a=hf=\dfrac{hc}{\lambda}=\Delta E\]

    \[\Delta E=3\mbox{.}43\cdot 10^{-19}J\]

But

    \[k_BT=1\mbox{.}38\cdot 10^{-23}J/K\cdot 300K=4\mbox{.}10^{-21}J\]

Now, according to the Boltzmann distribution, the population in any state is given by

    \[N_i=\exp\left(E_i/k_BT\right)\]

Therefore, the ration of N_a and N_b in the thermal equilibrium should be

    \[\dfrac{N_b(eq)}{N_a(eq)}=e^{-\frac{\Delta E}{k_BT}}=1\mbox{.}1\cdot 10^{-36}\]

and thus

    \[N_b(eq)=N_a(eq)\cdot 1\mbox{.}1\cdot 10^{-36}\]

Using the giving value of N_a, we get

    \[N_b(eq)=5\mbox{.}5\cdot 10^{-16}\]

and that is a very tiny number of atoms. This is showing to us that the energy gap between the given atomic states at room temperature is so large that almost all the electrons “choose” to stay in the lower state and hardly we will find electrons in the upper state.

Solution to 2). When we pump atoms into the upper state, we create a non-equilibrium atomic state. Energy will be released up until the equilibrium is restored (in fact, this is the working principle of the laser, in a simplified fashion!). We obtain now

    \[N_a(non-eq)=5\cdot 10^{20}\]

    \[N_b(non-eq)=9\cdot 10^{20}\]

    \[N(non-eq)=N_a(non-eq)+N_b(non-eq)=14\cdot 10^{20}\]

and we are interested in the number of atoms which restores the thermal equilibrium. Since the ratio should be

    \[\dfrac{N_a}{N_b}=1\mbox{.}1\cdot 10^{-36}\]

remains the same (the gap width should not change), we find

    \[N_a(non-eq)+N_b(non-eq)=14\cdot 10^{20}\]

    \[N_b(non_eq)(1+9\mbox{.}09\cdot 10^{35})=9\cdot 10^{20}\]

and then

    \[N_b(non-eq)=1\mbox{.}54\cdot 10^{-15}\]

The number of atoms that contribute to restore the thermal equilibrium is

    \[\Delta N=N(non-eq)-N_b(non-eq)\approx 9\cdot 10^{20}\]

The energy released in a single monochromatic pulse would be

    \[E=\Delta N\cdot \dfrac{hc}{\lambda}=9\cdot 10^{20}\cdot 3\mbox{.}43\cdot 10^{-19}\approx 309J\]

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