91Ó°ÊÓ

Altitude at which the carbon molecules undergo laser action is .

From figure,$E_0=0\mathrm{eV}$,$E_1=0.165\mathrm{eV}$,$E_2=0.289\mathrm{eV}$

The population inversion occurs between the energy levels${\mathrm{E}}_1$ and ${\mathrm{E}}_2$.

Consider the known data below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63×{10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242×{10}^{18}×6.63×{10}^{-34}\right)\mathrm{eV}.\mathrm{s} \\ =41.384×{10}^{-16}\mathrm{eV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} \mathrm{c}=3×{10}^8\mathrm{m}/\mathrm{s} \\ =\left(3×{10}^8×{10}^{12}\right)\mathrm{pm}/\mathrm{s} \\ =3×{10}^{20}\mathrm{pm}/\mathrm{s} \end{gathered}$$

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

For the excitation of molecules due to sunlight from the ground state to the 2nd excited state, helps to get the wavelength of the sunlight rays using Planck's relation and the given data in the figure. Now for the lasing action, the excitation is found to be from the first excited state to the second excited state, thus using Planck's relation, helps to find the wavelength. Now, comparing the wavelengths to the electromagnetic spectrum, it is found within the infrared range of 700 nm to 1 mm.

Formula:

The energy of the photon due to Planck’s relation,

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{λ}}$ ….. (1)

The lasing action for excitation of molecules due to sunlight is from${\mathrm{E}}_0=0\mathrm{eV}$ to${\mathrm{E}}_2=0.289\mathrm{eV}$ (from figure).

Thus, the wavelength of the sunlight that excites the molecules in the lasing action is given using the given data in equation (1) as follows:

$$\begin{gathered} \mathrm{λ}=\frac{\left(41.384×{10}^{-16}\mathrm{eV}.\mathrm{s}\right)\left(3×{10}^{20}\mathrm{pm}/\mathrm{s}\right)}{0.289\mathrm{eV}-0\mathrm{eV}} \\ =\frac{1240\mathrm{eV}.\mathrm{nm}}{0.289\mathrm{eV}} \\ =4.29×{10}^3\mathrm{nm} \\ =4.29\mathrm{μ³¾} \end{gathered}$$

Hence, the value of the wavelength is $4.29\mathrm{μ³¾}$.

Using the concept, we can see that the lasing action occurs within the energy levels${\mathrm{E}}_1=0.165\mathrm{eV}$ to${\mathrm{E}}_2=0.289\mathrm{eV}$ (from figure).

Thus, the wavelength at which lasing action occurs is given using the given data in equation (1) as follows:

$$\begin{gathered} \mathrm{λ}\text{'}=\frac{1240\mathrm{eV}.\mathrm{nm}}{0.289\mathrm{eV}-0.165\mathrm{eV}} \\ =1×{10}^4\mathrm{nm} \\ =10\mathrm{μ³¾} \end{gathered}$$

Hence, the value of the wavelength is $10\mathrm{μ³¾}$.

Both the wavelengths$\mathrm{λ}=4.29\mathrm{μ³¾}$ and$\mathrm{λ}\text{'}=10\mathrm{μ³¾}$ belong to the infrared region (700 nm to 1 mm) of the electromagnetic spectrum.

Hence, the region is found to the infrared region.