Question

a three level laser emits light of wavelength 550nm what is the ratio of upper level E2 to that of lower level E1 in laser transition at 300K

Answer 1

Answer:

Population inversion can be achieved by optical pumping in which the atoms in the ground state are excited to higher states by absorption of pump light. The electrons in the excited levels decay by non-radiative transitions to a metastable lev

Answer 2

Answer:

The ratio of the upper-level E₂ to the lower-level E₁, $\frac{N_{2} }{N_{1} }$ =  $1.07\times 10^{-38}$.

Explanation:

Given,

The wavelength of a three laser light, λ = $550nm$ = $550 \times 10^{-9}m$

The temperature at the laser transition, T = $300K$

The ratio of the upper-level E₂ to the lower-level E₁, $\frac{N_{2} }{N_{1} }$ =?

As we know,

• The ratio of the population of upper-level E₂ to lower-level E₁ is given by the formula:
• $\frac{N_{2} }{N_{1} }$ = $e^{\frac{-\Delta E}{KT} }$        -------equation (1)

Here, K = Boltazmann constant = $1.38\times 10^{-23} J/K$

• ΔE = energy difference = $\frac{hc}{\lambda}$

Here, h = Planck's constant = $6.6 \times 10^{-34} J-s$

And c = speed of light = $3\times 10^{8}m/s$

Thus, ΔE =  $\frac{hc}{\lambda}$ = $\frac{6.6\times 10^{-34 } \times 3\times 10^{8}}{550\times 10^{-9} }$ = $3.6 times 10^{-19}J$

Now, after putting the calculated value of ΔE in the equation (1), we get:

• $\frac{N_{2} }{N_{1} }$ = $e^{\frac{-3.6\times 10^{-19} }{1.38\times 10^{-23} \times 300} }$
• $\frac{N_{2} }{N_{1} }$ = $e^{-87.43}$

• $\frac{N_{2} }{N_{1} }$ = $1.07\times 10^{-38}$               ( $e^{-87.43}$ =  $1.07\times 10^{-38}$ )

Hence, the ratio of the upper-level E₂ to the lower-level E₁, $\frac{N_{2} }{N_{1} }$ =  $1.07\times 10^{-38}$.