A stellar object is emitting radiation at \(3.55 \mathrm{~mm}\). (a) What type of electromagnetic spectrum is this radiation? (b) If a detector is capturing \(3.2 \times 10^{8}\) photons per second at this wavelength, what is the total energy of the photons detected in one hour?

Using the given wavelength (3.55 mm), we can consult the electromagnetic spectrum toidentify the type of radiation. The common types of electromagnetic radiation, ordered by decreasing wavelength, are: - Radio waves - Microwaves - Infrared - Visible light - Ultraviolet - X-rays - Gamma rays A wavelength of 3.55 mm falls within the microwave range (1 mm to 100 cm). So, the radiation being emitted by the stellar object is microwave radiation.

We are given the detection rate (3.2 x 10^8 photons per second) and the wavelength (3.55 mm). First, let's convert the wavelength to meters: 3.55 mm = 3.55 x 10^-3 m. Now, we'll use the Planck's equation to find the energy of a single photon: \[E = hf\] Where E is the energy, h is the Planck's constant (\(6.63 \times 10^{-34} \mathrm{Js}\)), and f is the frequency. To find the frequency, use the speed of light (c) and the given wavelength (λ): \(c = f\lambda\) Rearranging to get the frequency: \(f = \frac{c}{\lambda}\) Now, plug in the values: \(f = \frac{3 \times 10^8 \mathrm{ms^{-1}}}{3.55 \times 10^{-3}\mathrm{m}} \approx 8.45 \times 10^{10} \mathrm{s^{-1}}\) Now, calculate the energy of a single photon: \(E = (6.63 \times 10^{-34} \mathrm{Js})(8.45 \times 10^{10} \mathrm{s^{-1}}) \approx 5.60 \times 10^{-23} \mathrm{J}\) Next, find the total energy of the detected photons per second by multiplying the energy of a single photon by the detection rate: \(E_\mathrm{total} = (5.60 \times 10^{-23} \mathrm{J})(3.2 \times 10^8 \mathrm{photons/s}) \approx 1.79 \times 10^{-14} \mathrm{J/s}\) The total energy detected per second is approximately \(1.79 \times 10^{-14}\) J/s.

Finally, we'll find the total energy detected in one hour by multiplying the energy per second by the number of seconds in an hour: \(E_\mathrm{1h} = (1.79 \times 10^{-14} \mathrm{J/s})(3600 \mathrm{s/h}) \approx 6.44 \times 10^{-11} \mathrm{J}\) The total energy of the photons detected in one hour is approximately \(6.44 \times 10^{-11}\) J. To summarize the answers: (a) The radiation is microwave radiation. (b) The total energy of the photons detected in one hour is approximately \(6.44 \times 10^{-11}\) J.