The population ratio of the two states in he-ne laser that produces light of wavelength 6000 A at 300 K is
Answers
Answer:
Population ratio is \frac{N_{2} }{N_{1}} =5.94 \times 10^{34}
N
1
N
2
=5.94×10
34
.
Given:
\lambdaλ = 6000 A = 6 ×10^{-7}10
−7
m
T = 300 K
To find:
Ratio of population = \frac{N_{2} }{N_{1} }
N
1
N
2
= ?
Formula used:
\frac{N_{2} }{N_{1} }
N
1
N
2
= e^{\frac{\Delta E}{kT} }e
kT
ΔE
\Delta E = \frac{hc}{\lambda}ΔE=
λ
hc
Solution:
Energy is given by,
\Delta E = \frac{hc}{\lambda}ΔE=
λ
hc
\Delta E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7} }ΔE=
6×10
−7
6.63×10
−34
×3×10
8
\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
×300
3.315×10
−19
\DeltaE = 80.07\DeltaE=80.07
Ratio of population = 1\frac{N_{2} }{N_{1} }1
N
1
N
2
= e^{\frac{\Delta E}{kT} }e
kT
ΔE
\frac{N_{2} }{N_{1}} =5.94 \times 10^{34}
N
1
N
2
=5.94×10
34
Population ratio is \frac{N_{2} }{N_{1}} =5.94 \times 10^{34}
N
1
N
2
=5.94×10
34
.