Question

Some of the most powerful lasers are based on the energy levels of neodymium in solids, such as glass, as shown in Figure 30.65.

(a) What average wavelength light can pump the neodymium into the levels above its metastable state?

(b) Verify that the 1.17 eV transition produces 1.06 µm radiation. Figure 30.65 Neodymium atoms in glass have these energy levels, one of which is metastable. The group of levels above the metastable state is convenient for achieving a population inversion, since photons of many different energies can be absorbed by atoms in the ground state.

Short answer

a) The average wavelength of light that may lift neodymium above its metastable state is 589.254 nm

b) As a result, a 1.17 eV transition may be shown to produce 1.06 μm radiation.

Step-by-step solution

Determine the formulas:

Consider the formula for the energy of X ray photons as follows:

$$\mathbf{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\lambda }$$

Here,

λ = Wavelength

E = Energy of the x-ray photons

h = Planck's constant

c = Speed of light

Consider the formula for the change in energy from one level to other as follows:

$$\mathbf{\Delta }\mathbf{E}\mathbf{=}{\mathbf{E}}_{\mathbf{i}}\mathbf{-}{\mathbf{E}}_{\mathbf{f}}$$

Here,

$$\mathrm{\Delta }E=$$Change in energy

$$\begin{array}{c}{E}_i=\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{E}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\\ E_f=\mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{E}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\end{array}$$

Calculate average wavelength light can pump the neodymium

a)

Calculation:

From energy level diagram,

$$\begin{array}{l}{E}_{\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{w}\mathrm{e}}=2.1\mathrm{e}\mathrm{V}\\ E_g=0\mathrm{e}\mathrm{V}\end{array}$$

Energy released during transition from ground state to above metastable state is calculated:

$$\mathrm{\Delta }E=\left(2.1\mathrm{e}\mathrm{V}\right)-\left(0\mathrm{e}\mathrm{V}\right)$$

$$\mathrm{\Delta }E=2.1\mathrm{e}\mathrm{V}$$

Now, wavelength emitted by laser is calculated using Planck's equation

Substituting the values in equation of energy and solve.

$$\begin{array}{c}E=\frac{hc}{\lambda }2.1\mathrm{e}\mathrm{V}\\ =\frac{\left(6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}\right)\left(3\times {10}^8\frac{\mathrm{m}}{\mathrm{s}\mathrm{e}\mathrm{c}}\right)}{\lambda }\end{array}$$

Solve further as:

$$\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}\right)\left(3\times {10}^8\frac{\mathrm{m}}{\mathrm{s}\mathrm{e}\mathrm{c}}\right)}{2.1\mathrm{e}\mathrm{V}}\\ \lambda =\frac{\left(6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}\right)\left(3\times {10}^8\frac{\mathrm{m}}{\mathrm{s}\mathrm{e}\mathrm{c}}\right)}{\left(2.1\mathrm{e}\mathrm{V}\right)\left(\frac{1.602\times {10}^{-19}\mathrm{J}}{1\mathrm{e}\mathrm{V}}\right)}\\ \lambda =\frac{\left(6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}\right)\left(3\times {10}^8\frac{\mathrm{m}}{\mathrm{s}\mathrm{e}\mathrm{c}}\right)}{3.36\times {10}^{-19}\mathrm{J}}\\ \lambda =589.254\mathrm{n}\mathrm{m}\end{array}$$

The average wavelength light can pump the neodymium into the levels above its metastable state is 589.254 nm.

Step 3:Verifying that the 1.17 eV transition produces 1.06 µm radiation.

b)

Substituting the values in equation of energy and solve.

$$\begin{array}{l}1.17\mathrm{e}\mathrm{V}=\frac{\left(6.626\times {10}^{-34}\mathrm{J}\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\mathrm{e}\mathrm{c}\right)}{\lambda }\\ \lambda =\frac{\left(6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}\right)\left(3\times {10}^8\frac{\mathrm{m}}{\mathrm{s}\mathrm{e}\mathrm{c}}\right)}{1.17\mathrm{e}\mathrm{V}}\\ \lambda =\frac{\left(6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}\right)\left(3\times {10}^8\frac{\mathrm{m}}{\mathrm{s}\mathrm{e}\mathrm{c}}\right)}{1.17\mathrm{e}\mathrm{V}\times \left(\frac{1.602\times {10}^{-19}\mathrm{J}}{1\mathrm{e}\mathrm{V}}\right)}\\ \lambda =1.06\mu \mathrm{m}\end{array}$$

Therefore, it can be demonstrated that a.17 eV transition yeilds λ= 1.06 μm radiation.