91影视

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.

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

$\mathrm{E}=\mathrm{鈩�}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

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

Substitute $1.05脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s}$for $\mathrm{鈩�}$, $480\text{鈥塏/尘}$for $k$and $3脳{10}^{鈭�26}\text{鈥塳驳}$for $\mathrm{m}$in the above equation.

$\begin{array}{c}\mathrm{E}=(1.05脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s})\sqrt{\frac{480\text{鈥塏/尘}}{3脳{10}^{鈭�26}\text{鈥塳驳}}}\\ =(1.05脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s})\sqrt{1.6脳{10}^{28}\text{鈥塏/办驳}鈰�\text{m}}\\ =(1.05脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s})\sqrt{1.6脳{10}^{28}\text{鈥塳驳}鈰�{\text{m/s}}^{\text{2}}\text{/kg}鈰�\text{m}脳\frac{1\text{鈥塳驳}鈰�{\text{m/s}}^{\text{2}}}{1\text{鈥塏}}}\\ =(1.05脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s})\sqrt{1.6脳{10}^{28}{\text{鈥塻}}^{\text{-2}}}\end{array}$

Hence, further as:

$\begin{array}{c}\mathrm{E}=(1.05脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s})\sqrt{1.6脳{10}^{28}{\text{鈥塻}}^{\text{-2}}}\\ =(1.05脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s})(1.26脳{10}^{14}{\text{鈥塻}}^{鈭�1})\\ =1.3脳{10}^{鈭�20}\text{鈥塉}脳\frac{1.6脳{10}^{鈭�19}\text{鈥塉}}{1\text{鈥塭痴}}\\ =2.12脳{10}^{鈭�39}\text{鈥塭痴}\end{array}$

Thus, the uniform energy spacing is $2.12脳{10}^{鈭�39}\text{鈥塭痴}$.

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

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

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

Substitute $2.12脳{10}^{鈭�39}\text{鈥塭痴}$for $\mathrm{E}$in the above equation.

${\mathrm{E}}_1=2.12脳{10}^{鈭�39}\text{鈥塭痴}$

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

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

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

Substitute $2.12脳{10}^{鈭�39}\text{鈥塭痴}$for $\mathrm{E}$in the above equation.

$\begin{array}{c}{\mathrm{E}}_2=2脳2.12脳{10}^{鈭�39}\text{鈥塭痴}\\ \text{=4.24}脳{10}^{鈭�39}\text{鈥塭痴}\end{array}$

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

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

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

Substitute$2.12脳{10}^{鈭�39}\text{鈥塭痴}$ for $E$in the above equation.

$\begin{array}{c}{\mathrm{E}}_3=3脳2.12脳{10}^{鈭�39}\text{鈥塭痴}\\ \text{=6.36}脳{10}^{鈭�39}\text{鈥塭痴}\end{array}$

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

Thus, the several photon energies are $2.12脳{10}^{鈭�39}\text{鈥塭痴}$,$4.24脳{10}^{鈭�39}\text{鈥塭痴}$ and$6.36脳{10}^{鈭�39}\text{鈥塭痴}$ respectively. The photons belong in the spectrum鈥檚 infrared part.