Question

According to the energy-time uncertainty principle, the lifetime$\mathbf{∆}\mathit{l}$of a state and the uncertainty$\mathbf{∆}\mathit{E}$in its energy are invertible proportional. Hydrogen's red spectral line is the result of an electron making a transition "downward" frum a quantum state whose lifetime is about${\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{s}$.

(a) What inherent uncertainty in the energy of the emitted person does this imply? (Note: Unfortunately. we might use the symbol for the energy difference-i.e., the energy of the photon-but here it means the uncertainly in that energy difference.)

(b) To what range in wavelength s does this correspond? (As noted in Exercise 2.57. the uncertainty principle is one contributor to the broadening of spectral lines.)

(c) Obtain a general formula relating $\mathbf{∆}\mathbf{\lambda }\mathbf{}\mathbf{to}\mathbf{}\mathbf{∆}\mathbf{t}$.

Short answer

$$\begin{gathered} \left(a\right)∆E\ge 5.27\times {10}^{-27}J \\ \left(b\right)1.14\times {10}^{-14}m \\ \left(c\right)∆\lambda ∆t\ge \frac{{\lambda }^2}{4\mathrm{\pi c}} \end{gathered}$$

Step-by-step solution

The Heisenberg uncertainty principle

The Heisenberg uncertainty principle relating an uncertainty in energy $\mathbf{∆}\mathit{E}$with an uncertainty in time $\mathbf{∆}\mathit{t}$are related as

$\mathbf{∆}\mathit{E}\mathbf{∆}\mathit{t}\mathbf{\ge }\frac{\sout{\mathbf{h}}}{\mathbf{2}}$

Here, $\sout{\mathbf{h}}$is reduced Planck's constant.

Energy E of a photon with wavelength $\mathit{\lambda }$is given by

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

Here, h is Planck's constant, and c is speed of light in vacuum.

 Determine the energy of the emitted person

(a)

As given Wavelength$\mathrm{\lambda }=656\mathrm{nm}=(656\mathrm{nm})\left(\frac{{10}^{-9}\mathrm{m}}{\mathrm{nm}}\right)=656\times {10}^{-9}\mathrm{m}$

The uncertainty in time $∆\mathrm{t}={10}^{-\mathrm{s}}\mathrm{s}$

All values, $\left({10}^{-8}\mathrm{s}\right)$for the uncertainty in time $∆t$, and $1.0546\times {10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{for}h$ is taken in the equation uncertainty in energy $∆E$

$$\begin{gathered} ∆\mathrm{E}∆\mathrm{t}\ge {∆}^{\text{'}}\mathrm{E} \\ ∆\mathrm{E}\ge \frac{\sout{\mathrm{h}}}{2∆\mathrm{t}} \\ ∆\mathrm{E}\ge \frac{(1.0546\times {10}^{-34}\mathrm{J}.\mathrm{s})}{2({10}^{-8}\mathrm{s})} \\ ∆\mathrm{E}\ge 5.273\times {10}^{-27}\mathrm{J} \end{gathered}$$

Therefore, uncertainty in energy for that quantum level $∆E\ge 5.27\times {10}^{-27}\mathrm{J}$.

 Step 2: determine the derivation on both sides of  E=hcλ

(b)

The photon’s energy comprises of$\lambda$

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

Take the derivation on both sides

$$\begin{gathered} dE=\frac{hc}{\lambda }d\lambda \\ -\frac{{\lambda }^2}{hc}dE=d\lambda \end{gathered}$$

As we are only interested in the magnitude of the change in wavelength, the absolute value

$d\lambda =\frac{{\lambda }^2}{hc}dE$

Take $6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{for}\mathrm{h},3\times {10}^8\mathrm{m}/\mathrm{s}\mathrm{for}c\mathrm{and}5.27\times {10}^{-27}\mathrm{J}\mathrm{for}dE$in the above equation to solve for

$$\begin{gathered} d\lambda =\frac{{\lambda }^2}{hc}dE \\ d\lambda =\left[\frac{{(6.56\times {10}^{-7}\mathrm{m})}^2}{(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})(3\times {10}^8\mathrm{m}/\mathrm{s})}\right]\left(5.27\times {10}^{-27}\mathrm{J}\right) \\ d\lambda =1.14\times {10}^{-14}\mathrm{m} \end{gathered}$$

Hence, the range of wavelengths for the 656 nm spectral line is $1.14\times {10}^{-14}\mathrm{m}$.

Find the relationship between the ∆λ and ∆t 

(c)

The absolute value of the change in wavelength

$d\lambda =\frac{{\lambda }^2}{hc}dE$

For finite change, $∆\lambda =\frac{{\lambda }^2}{hc}∆E$

Take the value of $\frac{hc}{{\lambda }^2}d\lambda \mathrm{for}∆E,\mathrm{and}\frac{h}{2\mathrm{\pi }}\mathrm{for}\sout{h}$in the equation $∆E∆t\ge \frac{\sout{h}}{2}$

$$\begin{gathered} \left(\frac{hc∆\lambda }{{\lambda }^2}\right)∆t\ge \frac{\left({\displaystyle \frac{h}{2\mathrm{\pi }}}\right)}{2} \\ \frac{c∆\lambda ∆t}{{\lambda }^2}\ge \frac{1}{4\mathrm{\pi }} \\ ∆\lambda ∆t\ge \frac{{\lambda }^2}{4\mathrm{\pi c}} \end{gathered}$$

Hence, the relationship between the $∆\lambda$and $∆t$is $∆\lambda ∆t\ge \frac{{\lambda }^2}{4\mathrm{\pi c}}$.