Question

Consider a vibrating molecule that behaves as a simple harmonic oscillator of mass ${\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{kg}$, spring constant 10³N/m and charge is +e , (a) Estimate the transition time from the first excited state to the ground state, assuming that it decays by electric dipole radiation. (b) What is the wavelength of the photon emitted?

Short answer

(a)The transition time from the first excited state to the ground state is 0.69 ms.

(b)The wavelength of the photon emitted is$1.9\times {10}^{-6}\mathrm{m}$.

Step-by-step solution

Formula used:

The expected value of r between the initial and final states will be

$\mathbf{\int }\mathbf{r\Psi }{\mathbf{*}}_{\mathbf{f}}\left(\overline{r}\right){\mathbf{\Psi }}_{\mathbf{i}}\left(\overline{r}\right)\mathbf{dV}$

Where,role="math" localid="1659762410552" ${\mathbf{\Psi }}_{\mathbf{f}}$would be n=0 state,${\mathbf{\Psi }}_{\mathbf{i}}$would be n=1 state,rwould be x , and dV would be dx.

You also know that,

${\mathbf{\Psi }}_{\mathbf{0}}\mathbf{=}{\left(\frac{b}{\sqrt{\pi }}\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{}{{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{b}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}}}^{\mathbf{}\mathbf{}}$

Also,

${\mathbf{\Psi }}_{\mathbf{1}}\mathbf{=}{\mathbf{\left(}\frac{\mathbf{b}}{\mathbf{2}\sqrt{\mathbf{\pi }}}\mathbf{\right)}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{}{{\mathbf{bxe}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{b}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}}}^{\mathbf{}\mathbf{}}$

Angular frequency of the radiation:

Now,

${\int }_{-\infty }^{+\infty }{\Psi }_0{\Psi }_{1}dx=\frac{\sqrt{2}{b}^2}{\pi }{\int }_{-\infty }^{+\infty }b{x}^2{e}^{-b^2{x}^2}dx$

The Gaussian integral evaluates to$\frac{\sqrt{\mathrm{\pi }}}{{\mathrm{b}}^3}$.

Hence, expected value of distance is,

$\mathrm{r}=\frac{1}{\mathrm{b}\sqrt{2\mathrm{\pi }}}$.

If$\mathrm{b}={\left(\frac{\mathrm{mk}}{{\mathrm{h}}^2}\right)}^{1/4}$

Hence, by putting the value of and solving, you get

$$\begin{gathered} \mathrm{r}=\frac{1}{{\left({\displaystyle \frac{\mathrm{mk}}{{\mathrm{h}}^2}}\right)}^{1/4}\sqrt{2\mathrm{\pi }}} \\ =\frac{1}{{\left({\displaystyle \frac{{10}^{-27}\times {10}^3}{{\left(1.055\times {10}^{-19}\mathrm{J}\right)}^2}}\right)}^{1/4}\sqrt{2\times 3.14}} \\ =4.1\times {10}^{-12}\mathrm{m} \end{gathered}$$

Now,

Also, the energy difference is,

$$\begin{gathered} {\mathrm{E}}_1-{\mathrm{E}}_0=\mathrm{h}\sqrt{\mathrm{k}/\mathrm{m}} \\ =1.055\times \times {10}^{-34}\sqrt{\frac{{10}^3}{{10}^{-27}}} \\ =1.055\times \times {10}^{-19}\mathrm{J} \end{gathered}$$

Hence, the radiation’s angular frequency will be

$$\begin{gathered} \mathrm{\omega }=\frac{1.055\times {10}^{-19}\mathrm{J}}{1.055\times {10}^{-34}\mathrm{J}.\mathrm{S}} \\ ={10}^{15}{\mathrm{s}}^{-1} \end{gathered}$$

(a) Transition time from the first excited state to the ground state

By using the data available in previous steps define the transition time as below.

$$\begin{gathered} \mathrm{t}=\frac{12\left(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right){\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^3\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}{{\left(6.6\times {10}^{-31}\mathrm{CM}\right)}^2{\left({10}^{15}{\mathrm{s}}^{-1}\right)}^3} \\ =0.69\mathrm{ms} \end{gathered}$$

(b) Wavelength of the photon emitted

As you know that the energy is,

$\mathbf{E}\mathbf{=}\frac{\mathbf{hc}}{\mathbf{\lambda }}$

Where, E is the Energy of photon, h Plank’s constant, and, $\mathit{\lambda }$ wavelength of emitted light.

By using the data of previous steps you can find the wavelength as

$$\begin{gathered} 1.055\times {10}^{-19}\mathrm{J}=\frac{\left(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\mathrm{\lambda }} \\ \mathrm{\lambda }=1.9\times {10}^{-6}\mathrm{m} \end{gathered}$$

Hence, wavelength of the emitted photon is $\mathrm{\lambda }=1.9\times {10}^{-6}\mathrm{m}$.