Question

What is the range of frequencies in a 1 ns pulse of

(a) 1060nmInfrared laser light and

(b) 100MHzRadio waves?

(c) For which is the "uncertainty" in frequency, relative to its approximate value, larger?

Short answer

(a). $\Delta f\ge 7.9\times {10}^7{s}^{-1}$ is the range of frequencies in a 1ns pulse of 1060nm infrared laser light.

(b). $\Delta f\ge 7.9\times {10}^7{s}^{-1}$is the range of frequency.

(c). For radio waves, the uncertainty in frequency relative to its approximate value is larger.

Step-by-step solution

Concept of Heisenberg's Uncertainty Principle and formula of the speed of light, frequency, wavelength

Using uncertainty principle$\mathbf{∆}\mathit{E}\mathbf{∆}\mathit{t}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{2}}$

Where,$\mathbf{∆}\mathit{E}$is the uncertainty in energy,$\mathbf{∆}\mathit{t}$is the uncertainty in time,his Planck’s

constant.

Formula used,

$\mathit{c}\mathbf{=}\mathit{\lambda }\mathit{f}$

where, is the speed of light, fis the frequency,$\mathit{\lambda }$is the wavelength.

Use Heisenberg’s Uncertainty Principle for calculation.

(a)

Time

$$\begin{gathered} ∆t=1ns \\ =1\times {10}^{-9}s \end{gathered}$$

Wavelength

$$\begin{gathered} \lambda =1060\mathrm{nm} \\ =1060\times {10}^{-9}\mathrm{m} \end{gathered}$$

Therefore,

$$\begin{gathered} ∆E∆t\ge \frac{h}{2} \\ ∆\left(hf\right)∆t\ge \frac{\left({\displaystyle \frac{h}{2\mathrm{\pi }}}\right)}{2} \\ h∆f∆t\ge \frac{h}{4\mathrm{\pi }} \\ ∆f∆t\ge \frac{1}{4\mathrm{\pi }} \\ ∆f\ge \frac{1}{4\mathrm{\pi }∆\mathrm{t}} \end{gathered}$$

Plugin the values

$$\begin{gathered} ∆f\ge \frac{1}{4\mathrm{\pi }\left(1\times {10}^{-9}\right)} \\ \ge 0.079\times {10}^9 \\ \ge 7.9\times {10}^7{s}^{-1} \end{gathered}$$

$∆f\ge 7.9\times {10}^7{s}^{-1}$is the range of frequencies in a 1ns pulse of 1060nm infrared laser light.

Use Heisenberg’s Uncertainty Principle for calculation.

(b)

Time

$$\begin{gathered} ∆t=1ns \\ ∆t=1\times {10}^{-9}s \end{gathered}$$

Frequency

$$\begin{gathered} \mathrm{f}=100\mathrm{MHz} \\ \mathrm{f}=100\times {10}^6\mathrm{Hz} \end{gathered}$$

Therefore,

$$\begin{gathered} ∆E∆t\ge \frac{h}{2} \\ ∆\left(hf\right)∆t\ge \frac{\left({\displaystyle \frac{h}{2\mathrm{\pi }}}\right)}{2} \\ h∆f∆t\frac{h}{4\mathrm{\pi }} \\ ∆f∆t\ge \frac{1}{4\mathrm{\pi }} \\ ∆f\ge \frac{1}{4\mathrm{\pi }∆\mathrm{t}} \end{gathered}$$

Plugin the values

$$\begin{gathered} ∆f\ge \frac{1}{4\mathrm{\pi }\left(1\times {10}^{-9}\right)} \\ \ge 0.079\times {10}^9 \\ \ge 7.9\times {10}^7{s}^{-1} \\ ∆f\ge 7.9\times {10}^7{s}^{-1} \end{gathered}$$

$∆\mathrm{f}\ge 7.9\times {10}^7{\mathrm{s}}^{-1}$ is the range in frequencies a 1ns pulse of 1060nm infrared laser light.

Use the formula  for calculation.

(c)

For infrared laser light.

Substitute the given value in the equation

$$\begin{gathered} f=\frac{c}{\lambda } \\ f=\frac{3\times {10}^8}{1.06\times {10}^{-6}} \\ f=2.83\times {10}^{14}{s}^{-1} \end{gathered}$$

The relative size of uncertainty in frequency to the approximate frequency can be find as:

$$\begin{gathered} \frac{\Delta f}{f}=\frac{7.96\times {10}^1{s}^{-1}}{2.83\times {10}^{14}{s}^{-1}} \\ \frac{∆f}{f}=2.81\times {10}^{-7} \end{gathered}$$

For radio waves

$$\begin{gathered} \frac{∆f}{f}=\frac{7.96\times {10}^{\text{'}}{s}^{-1}}{1\times {10}^8{s}^{-1}} \\ \frac{∆f}{f}=0.796 \end{gathered}$$

For radio waves, the uncertainty in frequency relative to its approximate value is $2.81\times {10}^{-7}$while for infrared laser light the uncertainty in frequency relative to its approximate value is 0.796.

Therefore, uncertainty in frequency relative to its approximate value is larger for radio waves.