Question

He-Ne Laser that produces light of wavelength 6328 Å at 27 °C then calculate (1)
Energy difference between upper and lower level (ii) Population of two states,
​

Answer 1

The energy of a photon is E=

λ

hc

where h= Planck's constant, c= velocity of light.

So, E=

632.8×10

−9

(6.62×10

−34

)(3×10

8

)

=3.14×10

−19

J=1.96eV where (1eV=1.6×10

−19

J)

Explanation:

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Answer 2

Given: λ is 6328Å at 27 °C.

To find: (i) energy difference between upper and lower level.

             (ii) population of two states

Step-by-step explanation:

Step 1 of 2

By Planck's equation $E=hv$.

The frequency can be written as a ratio of speed of light and wavelength.

Therefore the energy difference will be=

$=\frac{6.262*10^{-34} *3*10^{8} }{6328*10^{-10} } \\=0.003141*10^{-16}\\=3.141*10^{-19}$

The energy difference is $3.141*10^{-19}$.

Step 2 of 2

Population of 2 states is calculated by formula:

$\frac{N_{2} }{N_{1}} =e^{\frac{-(E_{2} -E_{1}) }{K_{B}T } }$

$K_{B} =1.38*10^{-23}$ boltzmann constant

$\frac{N_{2} }{N_{1}} =e^{\frac{-3.141*10^{-19} }{1.38*10^{-23}*300 } }$

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

The energy difference is $3.141*10^{-19}J$.

Population of two states $\frac{N_{2} }{N_{1}} =5.94*10^{34}$.