Question

Calculate the population ratio of two states in he - ne laser that produces light of wavelength 6000 å at 300 k

Answer 1

Answer:

Population ratio is $\frac{N_{2} }{N_{1}} =5.94 \times 10^{34}$.

Given:

$\lambda$ = 6000 A = 6 ×$10^{-7}$ m

T = 300 K

To find:

Ratio of population = $\frac{N_{2} }{N_{1} }$ = ?

Formula used:

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

$\Delta E = \frac{hc}{\lambda}$

Solution:

Energy is given by,

$\Delta E = \frac{hc}{\lambda}$

$\Delta E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7} }$

$\Delta E = 3.315 \times 10^{-19}$

$\DeltaE = \frac{3.315 \times 10^{-19}}{1.38 \times 10^{-23} \times 300 }$

$\DeltaE = 80.07$

Ratio of population = $1\frac{N_{2} }{N_{1} }$ =  $e^{\frac{\Delta E}{kT} }$

$\frac{N_{2} }{N_{1}} =5.94 \times 10^{34}$

Population ratio is $\frac{N_{2} }{N_{1}} =5.94 \times 10^{34}$.

Answer 2

Answer:

Check Attachment

Explanation: