Question

A flashlight beam produces \(2.5 \mathrm{~W}\) of electromagnetic radiation in a natrow beam. Although the light it produces is white (all visible wavelengths), make the simplifying assumption that the wavelength is \(550 \mathrm{~nm}\), the middle of the visible spectum. (a) How many photons per second emanate from the flashlight? (b) What force would the beam exent on a "perfect" mirror (i.e., one that reflects all light completely)?

Short answer

The number of photons emanating per second is \(6.93 \times 10^{18}\) photons and the force exerted by the light beam on a perfect mirror is \(1.66 \times 10^{-8} N\).

Step-by-step solution

Calculate the frequency of the light

First, the frequency \(f\) of the light can be calculated as \(f = c/ \lambda\), where \(c = 3.00 \times 10^8 ms^{-1}\) is the speed of light and \(\lambda = 550 \times 10^{-9} m\) is the provided wavelength in meters. This results in \(f = 5.45 \times 10^{14} Hz\). This represents the frequency of the light emitted by the flashlight.

Calculate the energy of each photon

The energy \(E\) of each photon can be calculated with the formula \(E = hf\), where \(h = 6.63 \times 10^{-34} Js\) is the Planck's constant. This results in \(E = 3.61 \times 10^{-19} Joules\). This is the energy of each photon.

Calculate the number of photons per second

The number of photons per second, which is the same as the power of the flashlight divided by the energy of each photon, results in \(N = 2.5 W / 3.61 \times 10^{-19} J = 6.93 \times 10^{18}\) photons/sec.

Calculate the momentum of each photon

Using the formula \(p=E/c\), the momentum \(p\) of each photon is \(p = E/c = 3.61 \times 10^{-19} J / 3.00 \times 10^8 ms^{-1} = 1.20 \times 10^{-27} kg.m/s\)

Calculate the total change in momentum

The total change in momentum is the number of photons/sec times the change in momentum of each photon. When a photon is reflected, it reverses direction and the magnitude of the change in momentum for each photon is \(2p\). So the total change in momentum per second or force is \(F = 2pN = 2 * 1.20 \times 10^{-27} kg.m/s * 6.93 \times 10^{18} s^{-1} = 1.66 \times 10^{-8} N\). The force exerted on the mirror by the light is calculated to be 1.66 x 10^-8 N.

Verification of the result

The result can be verified by confirming the units and ensuring the accuracy of calculations in each step. The result should show that the force exerted by the light on a mirror is incredibly small, reflecting the small momentum change of individual photons even when multiplied by the large number of photons in a flashlight beam.

Key concepts

Understanding Electromagnetic Radiation

Electromagnetic radiation is a form of energy that travels through space at the speed of light, which is approximately 299,792,458 meters per second (\( c \)). It encompasses a wide range of wavelengths including not only visible light but also radio waves, microwaves, infrared radiation, ultraviolet light, X-rays, and gamma rays.

When we think about a flashlight beam, we are picturing visible light, which is just a small portion of the electromagnetic spectrum. This light behaves both as a wave and as a particle, which leads to the concept of photons - the fundamental particles of light.

In the given exercise, students are asked to consider a beam of white light with a simplification that assumes a single wavelength of 550 nm, representative of the middle of the visible spectrum. By doing so, the calculations become manageable and enable us to delve into the characteristics of the photons that compose the light.

Photon Energy and its Calculation

The energy of a photon is directly related to its frequency (\( f \)) and is given by the famous equation \( E = hf \) where \( h \) represents Planck's constant and \( f \) is the frequency of the electromagnetic radiation.

Planck's constant (\( h \)), a fundamental constant in physics, has a value of approximately 6.62607015 \times 10^-34 joule-seconds (Js). This constant is crucial in quantum mechanics and connects the energy of photons to their frequencies.

In the step-by-step solution, the frequency was calculated using the speed of light and the given wavelength. Then, with the calculated frequency and Planck's constant, the energy of each photon was determined. Understanding this relationship is key because it shows how altering the frequency or wavelength of light changes the energy of the photons it emits — higher frequency (shorter wavelength) light has more energetic photons.

The Significance of Planck's Constant

Planck's constant has profound implications in quantum mechanics, setting the scale of quantum effects. Its role in the energy of photons is just one of its many applications; it also appears in the Heisenberg Uncertainty Principle and the quantization of energy levels in atoms.

Understanding Planck's constant is crucial for solving problems involving photons or quantum behaviors. It allows us to calculate the energy of a photon when the frequency is known, or vice versa. In the context of the textbook exercise, it was the bridge between the frequency of the light from the flashlight and the energy of the individual photons making up the light.

Students approaching this concept need to recognize that although its numerical value might seem inconsequential due to its extreme smallness, its role in physics is monumental. It describes a fundamental property of the universe that has no classical analog — the discrete or 'quantized' nature of energy at microscopic scales.