Question

Make a rough estimate of this uniform energy spacing in electron volts (where $\mathbf{1}\mathbf{eV}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\text{\hspace{0.17em}J}}$). You will need to make some rough estimates of atomic properties based on prior work. For comparison with the spacing of these vibrational energy states, note that the spacing between quantized energy levels for "electronic" states such as in atomic hydrogen is of the order of several electron volts.

(b) List several photon energies that would be emitted if a number of these vibrational energy levels were occupied due to collisional excitation. To what region of the spectrum (x-ray, visible, microwave, etc.) do these photons belong? (See Figure 8.1 at the beginning of the chapter.)

Short answer

a. The uniform energy spacing is $2.12\times {10}^{-39}\text{\hspace{0.17em}eV}$.

b. The several photon energies are $2.12\times {10}^{-39}\text{\hspace{0.17em}eV}$,$4.24\times {10}^{-39}\text{\hspace{0.17em}eV}$ and $6.36\times {10}^{-39}\text{\hspace{0.17em}eV}$respectively. The photons belong in the spectrum’s infrared part.

Step-by-step solution

Significance of the spring constant

The spring constant is described as the ratio of the force that is affected by the spring and the displacement caused by the spring. It mainly determines the stiffness of a spring.

(a) Determination of the uniform energy spacing

The equation of the uniform spacing of energy is expressed as:

$\mathrm{E}=\mathrm{\hslash }\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

Here, $\mathrm{E}$is the uniform spacing of energy, $\mathrm{\hslash }$is the Planck’s constant,$k$is the spring constant and $\mathrm{m}$is the mass of an HCl molecule.

Substitute $1.05\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$for $\mathrm{\hslash }$, $480\text{\hspace{0.17em}N/m}$for $k$and $3\times {10}^{-26}\text{\hspace{0.17em}kg}$for $\mathrm{m}$in the above equation.

$\begin{array}{c}\mathrm{E}=(1.05\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s})\sqrt{\frac{480\text{\hspace{0.17em}N/m}}{3\times {10}^{-26}\text{\hspace{0.17em}kg}}}\\ =(1.05\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s})\sqrt{1.6\times {10}^{28}\text{\hspace{0.17em}N/kg}\cdot \text{m}}\\ =(1.05\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s})\sqrt{1.6\times {10}^{28}\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\text{/kg}\cdot \text{m}\times \frac{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}}{1\text{\hspace{0.17em}N}}}\\ =(1.05\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s})\sqrt{1.6\times {10}^{28}{\text{\hspace{0.17em}s}}^{\text{-2}}}\end{array}$

Hence, further as:

$\begin{array}{c}\mathrm{E}=(1.05\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s})\sqrt{1.6\times {10}^{28}{\text{\hspace{0.17em}s}}^{\text{-2}}}\\ =(1.05\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s})(1.26\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1})\\ =1.3\times {10}^{-20}\text{\hspace{0.17em}J}\times \frac{1.6\times {10}^{-19}\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}eV}}\\ =2.12\times {10}^{-39}\text{\hspace{0.17em}eV}\end{array}$

Thus, the uniform energy spacing is $2.12\times {10}^{-39}\text{\hspace{0.17em}eV}$.

(b) Determination of the photon energies

The equation of the photon energy in the ground state is expressed as:

$\begin{array}{c}{\mathrm{E}}_1=1\times \mathrm{E}\\ =\mathrm{E}\end{array}$

Here, ${\mathrm{E}}_1$is the photon energy in the ground state.

Substitute $2.12\times {10}^{-39}\text{\hspace{0.17em}eV}$for $\mathrm{E}$in the above equation.

${\mathrm{E}}_1=2.12\times {10}^{-39}\text{\hspace{0.17em}eV}$

The equation of the photon energy in the first excited state is expressed as:

$\begin{array}{c}{\mathrm{E}}_2=2\times \mathrm{E}\\ =2\mathrm{E}\end{array}$

Here,${\mathrm{E}}_2$is the photon energy in the first excited state.

Substitute $2.12\times {10}^{-39}\text{\hspace{0.17em}eV}$for $\mathrm{E}$in the above equation.

$\begin{array}{c}{\mathrm{E}}_2=2\times 2.12\times {10}^{-39}\text{\hspace{0.17em}eV}\\ \text{=4.24}\times {10}^{-39}\text{\hspace{0.17em}eV}\end{array}$

The equation of the photon energy in the second excited state is expressed as:

$\begin{array}{c}{\mathrm{E}}_3=3\times \mathrm{E}\\ =3\mathrm{E}\end{array}$

Here,$E_3$is the photon energy in the second excited state.

Substitute$2.12\times {10}^{-39}\text{\hspace{0.17em}eV}$ for $E$in the above equation.

$\begin{array}{c}{\mathrm{E}}_3=3\times 2.12\times {10}^{-39}\text{\hspace{0.17em}eV}\\ \text{=6.36}\times {10}^{-39}\text{\hspace{0.17em}eV}\end{array}$

According to the diagram, it can be identified that the photons mainly belong to the spectrum’s infrared part.

Thus, the several photon energies are $2.12\times {10}^{-39}\text{\hspace{0.17em}eV}$,$4.24\times {10}^{-39}\text{\hspace{0.17em}eV}$ and$6.36\times {10}^{-39}\text{\hspace{0.17em}eV}$ respectively. The photons belong in the spectrum’s infrared part.