Question

A ruby laser produces radiation of wavelength $633\mathrm{nm}$ in pulses whose duration is $1.00\times {10}^{-9}\mathrm{s}$. (a) If the laser produces $0.376\mathrm{J}$ of energy per pulse, how many photons are produced in each pulse? (b) Calculate the power (in watts) delivered by the laser per pulse $(1\mathrm{W}=1\mathrm{J}/\mathrm{s}).$

Short answer

(a) Approximately $1.19\times {10}^{18}$ photons per pulse. (b) The power is $3.76\times {10}^8\mathrm{W}$.

Step-by-step solution

Calculate Energy of a Single Photon

First, find the energy of one photon using the formula: $E=\frac{hc}{\lambda }$. Here, $h$ is Planck's constant $6.63\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$, $c$ is the speed of light $3.00\times {10}^8\mathrm{m}/\mathrm{s}$, and $\lambda$ is the wavelength $633\mathrm{nm}=633\times {10}^{-9}\mathrm{m}$. Substituting in these values, you will find the energy per photon.

Calculate the Number of Photons Per Pulse

Use the total energy per pulse and the energy per photon to find the number of photons. The formula is: $\text{Number of photons}=\frac{\text{Total energy per pulse}}{\text{Energy per photon}}$. Substitute the total energy of $0.376\mathrm{J}$ and the energy calculated in Step 1.

Calculate Power Delivered by the Laser Per Pulse

Calculate the power delivered per pulse using the formula: $P=\frac{E}{t}$, where $E$ is the total energy per pulse $0.376\mathrm{J}$ and $t$ is the duration of the pulse $1.00\times {10}^{-9}\mathrm{s}$. Substitute these values to find the power in watts.

Key concepts

Energy of a Photon

The energy of a single photon can be calculated by using the formula: $$E=\frac{hc}{\lambda }$$This formula shows that the energy $E$ depends on Planck's constant $h$, the speed of light $c$, and the photon's wavelength $\lambda$. The energy is inversely proportional to the wavelength, meaning that photons with shorter wavelengths have more energy. In everyday experiences, this translates to why ultraviolet light, which has a shorter wavelength than visible light, causes sunburn more easily. To solve the original exercise, we first needed to calculate the energy of a photon emitted by a ruby laser using the given wavelength of $633\mathrm{nm}$. By substituting the known values into the formula, we obtain the energy for a single photon emitted by the laser. This step is crucial since it allows us to then figure out how many such photons are released in one pulse of the laser.

Planck's Constant

Planck's constant, represented by $h$, is a fundamental constant in physics, valued at approximately $6.63\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$. It plays a crucial role in the field of quantum mechanics.Planck's constant forms the link between the energy of a photon and the frequency of its electromagnetic wave. This constant is essential for the calculation of a photon's energy, bridging the gap between macroscopic and microscopic phenomena. Physicists rely on Planck's constant to describe how energy is quantized in quantum systems. In our exercise, we use $h$ within the formula for calculating photon energy, emphasizing its role in translating the wave properties of light into corresponding quantum energy levels. Understanding this constant helps in grasping how energy levels and wave properties coexist in the realm of quantum physics.

Speed of Light

The speed of light, denoted as $c$, is a universal constant with the value of approximately $3.00\times {10}^8\text{meters per second (m/s)}$. It is not just the speed at which light travels in a vacuum, but it also represents an upper limit for the speed at which information or matter can travel.When calculating the energy of a photon, the speed of light is a key part of the formula $E=\frac{hc}{\lambda }$, linking the photon's energy to its wavelength. In many calculations across physics, especially those regarding electromagnetic radiation, $c$ serves as a fundamental underpinning constant. Having such a large and unchanging value means that light doesn't vary much in speed, regardless of its medium, although the medium can slightly reduce this speed. This concept aids in making precise predictions in physics, as seen in both this exercise and in broader scientific explorations.

Wavelength

Wavelength, represented by $\lambda$, is the distance between consecutive peaks of a wave. In the context of electromagnetic radiation, like light, it determines the type of radiation and its energy. The equation $E=\frac{hc}{\lambda }$ directly involves wavelength, illustrating how energy is inversely related to $\lambda$. Thus, the shorter the wavelength, the higher the energy. In our provided exercise, the wavelength of $633\mathrm{nm}$ means we are dealing with a form of visible red light. Knowing the wavelength allows us to calculate energy levels of photons, as well as understand where this light sits on the electromagnetic spectrum. It is a vital parameter in assessing the properties and behaviors of waves, impacting everything from color perception in visible light to communication technologies relying on radio waves. By understanding wavelength, students can better appreciate the nature of different types of radiation, from gamma rays to long radio waves.