A solid laser has two states at 300K and 500K. If it emits light of wavelength 7000Å, then calculate the relative population at two different temperatures and compare the result.
Answers
Answer:
Answer:
Population ratio is \frac{N_{2} }{N_{1}} =5.94 \times 10^{34}N1N2=5.94×1034 .
Given:
\lambdaλ = 6000 A = 6 ×10^{-7}10−7 m
T = 300 K
To find:
Ratio of population = \frac{N_{2} }{N_{1} }N1N2 = ?
Formula used:
\frac{N_{2} }{N_{1} }N1N2 = e^{\frac{\Delta E}{kT} }ekTΔE
\Delta E = \frac{hc}{\lambda}ΔE=λhc
Solution:
Energy is given by,
\Delta E = \frac{hc}{\lambda}ΔE=λhc
\Delta E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7} }ΔE=6×10−76.63×10−34×3×108
\Delta E = 3.315 \times 10^{-19}ΔE=3.315×10−19
\DeltaE = \frac{3.315 \times 10^{-19}}{1.38 \times 10^{-23} \times 300 }\DeltaE=1.38×10−23×3003.315×10−19
\DeltaE = 80.07\DeltaE=80.07
Ratio of population = 1\frac{N_{2} }{N_{1} }1N1N2 = e^{\frac{\Delta E}{kT} }ekTΔE
\frac{N_{2} }{N_{1}} =5.94 \times 10^{34}N1N2=5.94×1034
Population ratio is \frac{N_{2} }{N_{1}} =5.94 \times 10^{34}N1N2=5.94×1034 .
Answer:
I hope this helps :)
