91影视

The energy of the emitted photon equals the energy difference between the two levels; And since the excited state is above the ground state by $螖E=2.58eV$, so the energy of the emitted photon is $E_{纬}=2.58eV$.

From equation 38.1, the energy of a photon of wavelength $\mathit{位}$is given by:

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{位}}$

Solving for$位$ and substituting for our value for $E_{纬}$, we get the wavelength of the emitted photon:

$$\begin{gathered} 位=\frac{hc}{E_{纬}} \\ 位=\frac{4.136脳{10}^{鈭�15}脳3脳{10}^8}{2.58} \\ 位=4.81脳{10}^{鈭�7}\mathrm{m} \\ \mathrm{位}=481\mathrm{nm} \end{gathered}$$

From equation 39.3, the uncertainty$\mathbf{鈭�}\mathbf{E}$ in the energy of a state that is occupied for a time$\mathbf{鈭�}\mathit{t}$ is given by$\mathit{螖}\mathit{E}\mathit{螖}\mathit{t}\mathbf{鈮�}\frac{\mathbf{h}}{\mathbf{2}}$

Substituting for$螖t=1.64脳{10}^{-7}s$ (the time during which the atom remains in its excited state), we get:

$$\begin{gathered} \mathrm{螖}E鈮�\frac{魔}{2\mathrm{螖}t} \\ \mathrm{螖}E鈮�\frac{1.055脳{10}^{鈭�34}}{2\left(1.64脳{10}^{鈭�7}\right)} \\ \mathrm{螖}E鈮�3.22脳{10}^{鈭�28}\mathrm{J} \end{gathered}$$

So, the minimum uncertainty in the energy of this level is:

$螖E=3.22脳{10}^{-28}J=2脳{10}^{-9}eV$

Rearranging equation (1),$E位=hc$

From calculus,$螖f(x,y)=y\frac{鈭�f}{鈭�x}+x\frac{鈭�f}{鈭�y}$ ; Applying this to the equation above, we get:

$位螖E+E螖位=0$

Rearranging,

$\begin{array}{r}位\mathrm{螖}E=鈭�E\mathrm{螖}位\\ \frac{\mathrm{螖}E}{E}=鈭�\frac{\mathrm{螖}位}{位}\\ 鈭�\left|\frac{\mathrm{螖}E}{E}\right|=\left|\frac{\mathrm{螖}位}{位}\right|\end{array}$

Solving this equation for $鈭�位$, we get:$鈭�位=\frac{鈭�E}{E}位$

Now, we plug our values for$鈭�E.E_{纬}$ and $位$, so we get the smallest possible uncertainty in the wavelength of the photon:

$\begin{array}{r}\mathrm{螖}位=\frac{2脳{10}^{鈭�9}}{2.58}\left(481\right)\\ \mathrm{螖}位=2.75脳{10}^{鈭�7}\mathrm{nm}\end{array}$