Question

figure 39-28 shows the energy-level diagram for a finite, one-dimensional energy well that contains an electron. The nonquantized region begins at ${\mathbf{E}}_{\mathbf{4}}\mathbf{=}\mathbf{450}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{eV}$. Figure 39-28b gives the absorption spectrum of the electron when it is in the ground state—it can absorb at the indicated wavelengths: ${\mathbf{\lambda }}_{\mathbf{a}}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{588}\mathbf{}\mathbf{nm}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{\lambda }}_{\mathbf{b}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{8437}$and for any wavelength less than ${\mathbf{\lambda }}_{\mathbf{c}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{9108}\mathbf{}\mathbf{nm}$ . What is the energy of the first excited state?

Short answer

The energy of the first excited state is 109 eV.

Step-by-step solution

Introduction

An electron is shown travelling rightward toward the trap, in a region with a voltage ${\mathbf{V}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{V}$, where it has a kinetic energy of 2.00 eV .

Concept:

The specific amount of energy contain by electrons at the different distances from the nucleus is known as energy level.

The expression for energy in terms of wavelength is,

$∆E=\frac{hc}{\lambda }$

Here, h is the plank’s constant, c is the velocity of the light, and $\lambda$is the wavelength.

Calculated the energy associated with as follows.

$∆E=\frac{hc}{\lambda }$

From the figure, the energy associated with ${\lambda }_c$must be equal to the difference between the higher energy state and lower energy state. This is because the photon which has the wavelength less than ${\lambda }_c$has sufficient energy to raise the electron from ground state to Non-quantized region.

$E_4=E_0=\frac{hc}{{\lambda }_c}$

Here, ${\mathrm{E}}_0$is the ground state energy.

Substitute localid="1661775989877" $450.0\mathrm{eV}\mathrm{for}{\mathrm{E}}_4,4.1357\times {10}^{-15}\mathrm{eV}.\mathrm{s}\mathrm{for}\mathrm{h},3\times {10}^8\mathrm{m}/\mathrm{s}\mathrm{for}\mathrm{c},\mathrm{and}2.9108\mathrm{nm}\mathrm{for}{\mathrm{\lambda }}_{\mathrm{c}}$in the above equation.

localid="1661776054908" role="math" $$\begin{gathered} (450.0\mathrm{eV})-{\mathrm{E}}_0=\frac{(4.1357\times {10}^{-15}\mathrm{eV}.\mathrm{s})(3\times {10}^8\mathrm{m}/\mathrm{s})}{(2.9108\mathrm{nm})\left({\displaystyle \frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}}\right)} \\ {\mathrm{E}}_0=23.76\mathrm{eV} \end{gathered}$$

Find the energy of the first excited state:

The first excited state energy is the sum of the ground state energy of the longest wavelength photon.

$E_1=E_0+E$ ….. (1)

The energy of the longest wavelength of the photon is,

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

Substitute $4.1357\times {10}^{-15}\mathrm{eV}.\mathrm{s}\mathrm{for}\mathrm{h},3\times {10}^8\mathrm{m}/\mathrm{s}\mathrm{for}\mathrm{c},\mathrm{and}14.588\mathrm{nm}\mathrm{for}{\mathrm{\lambda }}_{\mathrm{a}}$in the above equation.

$$\begin{gathered} \mathrm{E}=\frac{\left(4.1357\times {10}^{-15}\mathrm{eV}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{(14.588\mathrm{nm})\left({\displaystyle \frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}}\right)} \\ =85\mathrm{eV} \end{gathered}$$

Substitute 58 eV for E and 23.76 eV for $E_0$into equation (1).

role="math" localid="1661775888320" $$\begin{gathered} {\mathrm{E}}_1=85\mathrm{eV}+23.76\mathrm{eV} \\ =108.8\mathrm{eV} \\ \approx 109\mathrm{eV} \end{gathered}$$

Hence, the energy of the first excited state is 109 eV .