Question

A radar pulse has an average wavelength of \(1.0 \mathrm{cm}\) and lasts for \(0.10 \mu \mathrm{s}\). (a) What is the average energy of the photons? (b) Approximately what is the least possible uncertainty in the energy of the photons?

Short answer

Answer: The average energy of the photons is 1.99 x 10^-23 J, and the least possible uncertainty in the energy of the photons is approximately 5.28 x 10^-28 J.

Step-by-step solution

Calculate the Energy of the Photons

To calculate the average energy of the photons, we can use the Planck-Einstein relation, which states that the energy (E) of a photon is related to its frequency (f) or wavelength (λ) by the formula: $$ E = h \cdot f = \frac{h \cdot c}{\lambda} $$ where h is the Planck's constant \((6.626 \times 10^{-34} \mathrm{J \cdot s})\), c is the speed of light \((3.0 \times 10^8 \, \mathrm{m/s})\), and λ is the wavelength of the photons (given as \(1.0\,\mathrm{cm}\)). First, we need to convert the wavelength from cm to meters: $$ \lambda = 1.0 \, \mathrm{cm} \times \frac{1 \, \mathrm{m}}{100 \, \mathrm{cm}} = 0.01 \, \mathrm{m} $$ Now, we can plug the values into the formula and calculate the energy: $$ E = \frac{(6.626 \times 10^{-34} \, \mathrm{J \cdot s}) \cdot (3.0 \times 10^8 \, \mathrm{m/s})}{0.01 \, \mathrm{m}} = 1.99 \times 10^{-23} \, \mathrm{J} $$

Calculate the Least Possible Uncertainty in the Energy of the Photons

To find the least possible uncertainty in the energy of the photons, we can use the time-energy uncertainty principle, which states: $$ \Delta E \cdot \Delta t \geq \frac{h}{4 \pi} $$ where ΔE is the uncertainty in the energy of the photons, Δt is the uncertainty in time (given as \(0.10\, \mu\mathrm{s}\)), and h is the Planck's constant. First, we need to convert the uncertainty in time from microseconds to seconds: $$ \Delta t = 0.10 \, \mu\mathrm{s} \times \frac{1 \, \mathrm{s}}{10^6 \, \mu\mathrm{s}} = 1.0 \times 10^{-7} \, \mathrm{s} $$ Now, we can rearrange the time-energy uncertainty principle equation to find the least possible uncertainty in the energy (ΔE) of the photons: $$ \Delta E \geq \frac{h}{4 \pi \cdot \Delta t} $$ Finally, plug the values into the formula and calculate the least possible uncertainty in the energy: $$ \Delta E \geq \frac{(6.626 \times 10^{-34} \, \mathrm{J \cdot s})}{4 \pi \cdot (1.0 \times 10^{-7} \, \mathrm{s})} = 5.28 \times 10^{-28} \, \mathrm{J} $$ So the answers are (a) The average energy of the photons is \(1.99 \times 10^{-23} \, \mathrm{J}\), and (b) the least possible uncertainty in the energy of the photons is approximately \(5.28 \times 10^{-28} \, \mathrm{J}\).