Question

find the ratio of population of two energy levels in a laser if the transition between them produces light of wavelength 694.3 nm. assume the ambient temperature to be 27 degree celcius

Answer 1

Explanation:

This is the solution. I've attached a photo

Answer 2

Answer:

The required ratio is $2.01\times 10^{-30}$.

Explanation:

• The relative population is controlled by the difference in energy levels between the ground state and the system's temperature.
• A high value of energy difference leads to a low population in the higher energy state.

Given:

Wavelength, λ = 694.3 nm = $694.3\times 10^{-9} m$

Temperature, T = 27° celcius = $273+27=300K$

To Find:

The ratio of Population = $\frac{N_{1} }{N_{2} }$

Formula to be used:

                                  $\frac{N_{1} }{N_{2} } =e^{-\frac{\delta E}{kT} }$

Step-1:

Find $\delta E=\frac{hc}{\lambda}$

Here,

h = $6.63 \times 10^{-34}$

c = $3\times 10^{8}$

      $\delta E =\frac{6.63\times 10^{-34}\times 3 \times 10^{8} }{694.3\times 10^{-9} }$$= 2.86 \times 10^{-19}$

Step-2:

Find $\frac{\delta E}{kT}$:

where,

k = $1.38\times 10^{-23}$

           $\frac{\delta E}{kT}=\frac{2.83 \times 10^{-19} }{1.38 \times 10^{-23}\times 300 }$$= 68.35$

Step-3:

           $\frac{N_{1} }{N_{2} } = e^{-68.35} =2.01 \times 10^{-30}$

Thus, the ratio of the population of two energy levels is $2.01\times 10^{-30}$.