Problem 31 - Chemistry: The Central Science

Question

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?

Step-by-Step Solution

Answer

(a) The radiation belongs to the near-infrared region of the electromagnetic spectrum. (b) The diode laser emits approximately \(8.04 \times 10^{16}\) photons per second.

Step 1: Identify the portion of the electromagnetic spectrum

The laser emits at a wavelength of \(987 nm\), which is equivalent to \(987 \times 10^{-9} m\). By referring to the electromagnetic spectrum, we can find that this wavelength belongs to the near-infrared region.

Step 2: Calculate the energy of a single photon

First, we need to find the frequency of the emitted radiation using the speed of light equation: \(c = \nu \lambda \Rightarrow \nu = \frac{c}{\lambda}\) Plugging in the values, we get: \(\nu = \frac{3 \times 10^8 m/s}{987 \times 10^{-9} m} \approx 3.04 \times 10^{14} Hz\) Now, we can calculate the energy of a single photon using the energy of a photon equation: \(E = h\nu\) Plugging in the values, we get: \(E = (6.63 \times 10^{-34} Js)(3.04 \times 10^{14} Hz) \approx 2.02 \times 10^{-19} J\)

Step 3: Calculate the number of photons emitted per second

To find the number of photons emitted per second, we need to use the given total energy and period: Total energy absorbed = \(0.52 J\) Period = \(32 s\) First, we find the energy absorbed per second (also known as the power): Power = \(\frac{Total~Energy}{Period}\) Power = \( \frac{0.52 J}{32 s} \approx 0.01625 J/s \) Now, we divide the energy absorbed per second by the energy of a single photon to find the number of photons emitted per second: Number of photons per second = \(\frac{Power}{Energy~per~photon}\) Number of photons per second = \( \frac{0.01625 J/s}{2.02 \times 10^{-19} J} \approx 8.04 \times 10^{16}~photon/s\) Therefore, the diode laser is emitting approximately \(8.04 \times 10^{16}\) photons per second.