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?

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: $$ E = \frac{(6.626 \times 10^{-34} \, \mathrm{J \cdot s}) \cdot (3.0 \times 10^8 \, \mathrm{m/s})}{0.01 \, \mathrm{m}} = 1.99 \times 10^{-23} \, \mathrm{J} $$

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 \(5.28 \times 10^{-28} \, \mathrm{J}\).