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?

Short answer

The radiation emitted by the diode laser is found in the infrared portion of the electromagnetic spectrum. The laser emits approximately \(8.06 \times 10^{16}\) photons per second.

Step-by-step solution

Part (a): Identifying the portion of electromagnetic spectrum

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.

Part (b): Calculating the number of photons emitted per second

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.

Key concepts

Infrared Radiation

Infrared radiation is a type of electromagnetic radiation that is not visible to the human eye but can be felt as heat. It has longer wavelengths than visible light, ranging from approximately 700 nanometers (nm) to 1 millimeter (mm). This positions it between visible light and microwaves on the electromagnetic spectrum.

In the context of the exercise, the given wavelength of the diode laser is 987 nm, which falls within the infrared range. Infrared radiation is commonly used in various applications, such as remote controls, thermal imaging, and also in laser technology for communication and sensing. Understanding where a particular wavelength falls in the electromagnetic spectrum helps in determining suitable applications and understanding the properties of that radiation.

Photon Energy Calculation

The energy of a photon is a crucial concept in understanding how many photons are emitted by a light source, such as a diode laser. The energy of individual photons is calculated using the equation: \[E_{photon} = \frac{hc}{\lambda}\] where:

• \(E_{photon}\) stands for the energy of a photon.
• \(h\) is Planck's constant, approximately \(6.63 \times 10^{-34}\) Joules per second (J·s).
• \(c\) is the speed of light, approximately \(3.00 \times 10^8\) meters per second (m/s).
• \(\lambda\) is the wavelength of the light in meters.

By substituting the values for a wavelength of 987 nm (converted to meters as \(9.87 \times 10^{-7}\) m), we determine that the energy per photon of this infrared radiation is about \(2.019 \times 10^{-19}\) Joules.

This concept of photon energy is essential for calculating how much energy is emitted over time and directly applies to other areas like energy efficiency and quantum physics.

Diode Laser Emission

A diode laser is a type of laser device that uses a semiconductor to produce light. These lasers are popular in electronics and communication due to their efficiency and compact size. Diode lasers emit coherent light of a specific wavelength, which, in the exercise, is in the infrared region.

To determine how many photons a diode laser emits per second, one must account for the total energy output over a set period and the energy of individual photons, as earlier discussed. By dividing the total energy output over 32 seconds (0.52 Joules) by the energy per photon (\(2.019 \times 10^{-19}\) Joules), the total number of emitted photons can be calculated.

Finally, dividing the total number of photons by time gives the emission rate: approximately \(8.06 \times 10^{16}\) photons per second. This calculation demonstrates the significant output possible even from small laser devices, showcasing their potent application in technology and medicine where precision radiation is crucial.