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The metastable state of a laser active medium with 3 energy levels is 1.4 eVabove the non-excited level at 0eV. If the non-radiative transition from the excitedstate to metastable state is 1.15μm, find out(a) Position of excited state in eV.(b) Laser photon wavelength(c) Wavelength of the pumping source

Question

The metastable state of a laser active medium with 3 energy levels is 1.4 eVabove the non-excited level at 0eV. If the non-radiative transition from the excitedstate to metastable state is 1.15μm, find out(a) Position of excited state in eV.(b) Laser photon wavelength(c) Wavelength of the pumping source

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Solution

(a) Position of excited state in eV

The energy of the non-radiative transition from the excited state to the metastable state can be calculated using the formula:

E = hc/λ

where: E is the energy, h is Planck's constant (6.626 x 10^-34 J.s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength (1.15 x 10^-6 m).

Substituting the values into the formula, we get:

E = (6.626 x 10^-34 J.s)(3 x 10^8 m/s) / (1.15 x 10^-6 m)

E = 1.73 x 10^-19 J

To convert this energy to electron volts (eV), we use the conversion factor 1 eV = 1.6 x 10^-19 J. Therefore,

E = 1.73 x 10^-19 J / 1.6 x 10^-19 J/eV

E = 1.08 eV

The position of the excited state is the sum of the energy of the metastable state and the energy of the non-radiative transition. Therefore,

Position of excited state = 1.4 eV + 1.08 eV = 2.48 eV

(b) Laser photon wavelength

The energy of the laser photon is the difference between the energy of the metastable state and the non-excited state, which is 1.4 eV. To find the wavelength of the laser photon, we can use the formula:

λ = hc/E

Substituting the values into the formula, we get:

λ = (6.626 x 10^-34 J.s)(3 x 10^8 m/s) / (1.4 eV)(1.6 x 10^-19 J/eV)

λ = 8.85 x 10^-7 m = 885 nm

(c) Wavelength of the pumping source

The energy of the pumping source is the same as the energy of the excited state, which is 2.48 eV. To find the wavelength of the pumping source, we can use the formula:

λ = hc/E

Substituting the values into the formula, we get:

λ = (6.626 x 10^-34 J.s)(3 x 10^8 m/s) / (2.48 eV)(1.6 x 10^-19 J/eV)

λ = 5.01 x 10^-7 m = 501 nm

This problem has been solved

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