A diode laser emits at a wavelength of \(987 \mathrm{nm} .\) (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of $0.52 \mathrm{~J}\( over a period of \)32 \mathrm{~s}$. How many photons per second are being emitted by the laser?

The given wavelength of the radiation is \(987 \mathrm{nm}\), which is \(9.87 \times 10^{-7} \mathrm{m}\). We need to compare this value with the ranges of different portions of the electromagnetic spectrum to determine where this radiation lies. Here are the approximate wavelength ranges for different portions of the electromagnetic spectrum: - Radio waves: > 1000 m - Microwaves: 1 mm - 1 m - Infrared: 700 nm - 1 mm - Visible light: 400 nm - 700 nm - Ultraviolet: 10 nm - 400 nm - X-rays: 0.01 nm - 10 nm - Gamma rays: < 0.01 nm Since the given wavelength (987 nm) falls within the range of the infrared spectrum (700 nm to 1 mm), this radiation is found in the infrared portion of the electromagnetic spectrum.

To determine the number of photons emitted per second, we first need to calculate the energy of a single photon. The energy of a photon can be calculated using the following formula: \[E_{photon} = \dfrac{hc}{\lambda}\] Where \(E_{photon}\) is the energy of a photon, \(h\) is the Planck's constant (\(6.63 \times 10^{-34} \mathrm{J \cdot s}\)), \(c\) is the speed of light (\(3.00 \times 10^{8} \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the light (\(9.87 \times 10^{-7} \mathrm{m}\)). Now, let's calculate the energy of a single photon: \[E_{photon} = \dfrac{(6.63 \times 10^{-34} \mathrm{J \cdot s}) (3.00 \times 10^{8} \mathrm{m/s})}{9.87 \times 10^{-7} \mathrm{m}}\] \[E_{photon} ≈ 2.019 \times 10^{-19} \mathrm{J}\] Now that we know the energy of a single photon, we can use the given output energy (\(0.52\, \mathrm{J}\)) and the total time (\(32 \, \mathrm{s}\)) to find the number of photons emitted per second. First, let's find the total number of photons emitted: \[N_{total} = \dfrac{E_{total}}{E_{photon}}\] \[N_{total} = \dfrac{0.52 \, \mathrm{J}}{2.019 \times 10^{-19} \, \mathrm{J}}\] \[N_{total} ≈ 2.58 \times 10^{18}\] Now, we can find the number of photons emitted per second: \[N_{per \, second} = \dfrac{N_{total}}{t}\] \[N_{per \, second} = \dfrac{2.58 \times 10^{18}}{32 \, \mathrm{s}}\] \[N_{per \, second} ≈ 8.06 \times 10^{16}\] Therefore, the laser emits approximately \(8.06 \times 10^{16}\) photons per second.