91影视

The given data can be listed below as:

• The energy at the ground state is ${\mathrm{E}}_1=鈭�2.3\text{鈥塭痴}$.
• The energy at the first excited state is ${\mathrm{E}}_2=鈭�1.1\text{鈥塭痴}$.
• The energy at the second excited state is ${\mathrm{E}}_3=鈭�0.5\text{鈥塭痴}$.
• The energy at the third excited state is ${\mathrm{E}}_4=鈭�0.2\text{鈥塭痴}$.

The energy graph is described as a graph in which an energy change occurs while a particular reaction occurs. It mainly includes transition, reactants and product states.

The equation of the force is expressed as:

$\mathrm{F}=鈭�\mathrm{螖鲍}$

Here,$\mathrm{F}$ is the force and $\mathrm{螖鲍}$is the change in the potential energy. As the force increases, the potential energy slope also increases.

The negative largest slope is being contained by the point A. The force is being cancelled out as the slope has a negative sign and it moves in the$+\mathrm{r}$ direction. Having a slope of $0$, the force also becomes$0$ and the point B stays in the equilibrium position. The positive slope is being contained by the point C and the slope鈥檚 magnitude of point C is lesser than the slope鈥檚 magnitude of point A. As the point C has a positive slope and also a negative sign, it moves in the $鈭�\mathrm{r}$direction.

The diagram of the height of the kinetic energy when the system is at the location D has been drawn below:

In the above diagram, the red line is the line which adds up. The total energy is the summation of the kinetic and the potential energy.

Thus, the direction of the force is largest at the point A and it is at the$+\mathrm{r}$ direction. The direction of the force is the second largest at the point C and it is at the$-r$ direction. The direction of the force is smallest at the point B as it is $0$.

The particle can go to a distance of $\mathrm{r}=\mathrm{鈭�}$if it is set free. In this case, the particle has zero energy who can go to that distance. That shows that the particle needs a large amount of kinetic energy from going to the excited state from the ground state.

The equation of the required energy is expressed as:

$\begin{array}{c}\mathrm{k}+{\mathrm{E}}_1=0\\ \mathrm{K}=鈭�{\mathrm{E}}_1\end{array}$

Here,$K$ is the required energy and${\mathrm{E}}_1$ is the potential energy.

As the lowest energy according to the diagram in the question is $鈭�2.3\text{鈥塭痴}$, then it will be considered as the potential energy.

Substitute$鈭�2.3\text{鈥塭痴}$ for ${\mathrm{E}}_1$in the above equation.

$\begin{array}{c}K=鈭�(鈭�2.3\text{鈥塭痴})\\ =2.3\text{鈥塭痴}\end{array}$

Thus, the energy required to break the molecule apart is$2.3\text{鈥塭痴}$ .

The equation of the energy of the emitted light in the first phase is expressed as:

${\mathrm{P}}_1={\mathrm{E}}_2鈭�{\mathrm{E}}_1$

Here,${\mathrm{P}}_1$is the energy of the emitted light in the first phase,${\mathrm{E}}_2$is the energy at the first excited state and${\mathrm{E}}_1$is the energy at the ground state.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{P}}_1=\text{-1.1鈥塭痴-}\left(\text{-2.3鈥塭痴}\right)\\ =1.2\text{鈥塭痴}\end{array}$

The equation of the energy of the emitted light in the second phase is expressed as:

${\mathrm{P}}_2={\mathrm{E}}_3鈭�{\mathrm{E}}_1$

Here, ${\mathrm{P}}_2$is the energy of the emitted light in the second phase, ${\mathrm{E}}_3$is the energy at the second excited state and ${\mathrm{E}}_1$is the energy at the ground state.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{P}}_2=\text{-0.5鈥塭痴-}\left(\text{-2.3鈥塭痴}\right)\\ =1.8\text{鈥塭痴}\end{array}$

The equation of the energy of the emitted light in the third phase is expressed as:

${\mathrm{P}}_3={\mathrm{E}}_4鈭�{\mathrm{E}}_1$

Here, ${\mathrm{P}}_3$is the energy of the emitted light in the third phase, ${\mathrm{E}}_4$is the energy at the third excited state and ${\mathrm{E}}_1$is the energy at the ground state.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{P}}_3=\text{-0.2鈥塭痴-}\left(\text{-2.3鈥塭痴}\right)\\ =2.1\text{鈥塭痴}\end{array}$

The equation of the energy of the emitted light in the fourth phase is expressed as:

${\mathrm{P}}_4={\mathrm{E}}_3鈭�{\mathrm{E}}_2$

Here, ${\mathrm{P}}_4$is the energy of the emitted light in the fourth phase, ${\mathrm{E}}_3$is the energy at the second excited state and ${\mathrm{E}}_2$is the energy at the first excited state.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{P}}_4=\text{-0.5鈥塭痴-}\left(\text{-1.1鈥塭痴}\right)\\ =0.6\text{鈥塭痴}\end{array}$

The equation of the energy of the emitted light in the fifth phase is expressed as:

${\mathrm{P}}_5={\mathrm{E}}_4鈭�{\mathrm{E}}_2$

Here,${\mathrm{P}}_5$is the energy of the emitted light in the fifth phase,${\mathrm{E}}_2$is the energy at the first excited state and${\mathrm{E}}_4$is the energy at the third excited state.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{P}}_5=\text{-0.2鈥塭痴-}\left(\text{-1.1鈥塭痴}\right)\\ =0.9\text{鈥塭痴}\end{array}$

The equation of the energy of the emitted light in the sixth phase is expressed as:

${\mathrm{P}}_6={\mathrm{E}}_4鈭�{\mathrm{E}}_3$

Here, ${\mathrm{P}}_6$is the energy of the emitted light in the sixth phase, ${\mathrm{E}}_3$is the energy at the second excited state and ${\mathrm{E}}_4$is the energy at the third excited state.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{P}}_6=\text{-0.2鈥塭痴-}\left(\text{-0.5鈥塭痴}\right)\\ =0.3\text{鈥塭痴}\end{array}$

Thus, the energies in electron volts of the emitted light are $1.2\mathrm{eV},1.8\mathrm{eV},2.1\mathrm{eV},0.6\mathrm{eV},0.9\mathrm{eV}$and $0.3\mathrm{eV}$respectively.

The equation of the inertial mass of the molecule in the ground state is expressed as:

$\begin{array}{c}\mathrm{螖贰}={\mathrm{mc}}^2\\ \mathrm{m}=\frac{\mathrm{螖贰}}{{\mathrm{c}}^2}\end{array}$

Here,$\mathrm{m}$is the inertial mass of the molecule, $\mathrm{螖贰}$is the change of the energy states and$\mathrm{c}$is the velocity of light.

For the ground state,${\mathrm{m}}_1$is the mass.

Substitute $2.3\text{鈥塭痴}$ for $\mathrm{螖贰}$and $3脳{10}^8\text{鈥尘/蝉}$for $\mathrm{c}$ in the above equation.

$\begin{array}{c}{\mathrm{m}}_1=\frac{2.3\text{鈥塭痴}}{{(3脳{10}^8\text{鈥尘/蝉})}^2}\\ =\frac{2.3\text{鈥塭痴}脳\frac{1.6脳{10}^{鈭�19}\text{鈥塉}}{1\text{鈥塭痴}}}{(9脳{10}^{16}{\text{鈥尘}}^2{\text{/s}}^2)}\\ =\frac{3.6脳{10}^{鈭�19}\text{鈥塉}脳\frac{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{1\text{鈥塉}}}{(9脳{10}^{16}{\text{鈥尘}}^2{\text{/s}}^2)}\\ =4.08脳{10}^{鈭�36}\text{鈥塳驳}\end{array}$

For the excited state ,$m_2$is the mass.

Substitute $10\text{鈥塭痴}鈭�2.3\text{鈥塭痴}$ for $\mathrm{螖贰}$ and $3脳{10}^8\text{鈥尘/蝉}$for $\mathrm{c}$in the above equation.

$\begin{array}{c}{\mathrm{m}}_2=\frac{10\text{鈥塭痴}鈭�2.3\text{鈥塭痴}}{{(3脳{10}^8\text{鈥尘/蝉})}^2}\\ =\frac{7.7\text{鈥塭痴}脳\frac{1.6脳{10}^{鈭�19}\text{鈥塉}}{1\text{鈥塭痴}}}{(9脳{10}^{16}{\text{鈥尘}}^2{\text{/s}}^2)}\\ =\frac{1.2脳{10}^{鈭�18}\text{鈥塉}脳\frac{1\text{鈥塳驳}鈰�{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{1\text{鈥塉}}}{(9脳{10}^{16}{\text{鈥尘}}^2{\text{/s}}^2)}\\ =2.19脳{10}^{鈭�35}\text{鈥塳驳}\end{array}$

The equation of the difference between the mass of the ground and the excited state is expressed as:

$\mathrm{m}={\mathrm{m}}_2鈭�{\mathrm{m}}_1$

Here, $\mathrm{m}$is the difference between the mass of the ground and the excited state, ${\mathrm{m}}_1$is the mass of the ground state and ${\mathrm{m}}_2$is the mass of the excited state.

Substitute the values in the above equation.

$\begin{array}{c}\mathrm{m}=2.19脳{10}^{鈭�35}\text{鈥塳驳-}4.08脳{10}^{鈭�36}\text{鈥塳驳}\\ \text{=1.7}脳{10}^{鈭�35}\text{鈥塳驳}\end{array}$

Thus, the inertial mass in the ground state is higher by about$\text{1.7}脳{10}^{鈭�35}\text{鈥塳驳}$ than the mass in the excited state. The reason for the difference is due to the difference in the energy levels.