Problem 25 - University Physics with Modern Physics

Question

A laser produces light of wavelength 625 nm in an ultrashort pulse. What is the minimum duration of the pulse if the minimum uncertainty in the energy of the photons is 1.0\(\%\)?

Step-by-Step Solution

Answer

The minimum pulse duration is approximately \( 1.66 \times 10^{-14} \text{ s} \).

Step 1: Determine Energy of a Photon

First, we need to calculate the energy of a single photon using the given wavelength of 625 nm. The formula to find the energy of a photon is \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J s} \) (Planck's constant), \( c = 3 \times 10^8 \text{ m/s} \) (speed of light), and \( \lambda = 625 \times 10^{-9} \text{ m} \). Substitute these values into the formula to find \( E \).

Step 2: Calculate Exact Energy of Photon

Substitute the given values into the photon energy formula: \[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{625 \times 10^{-9}} \] Calculating this yields \( E \approx 3.18 \times 10^{-19} \text{ J} \).

Step 3: Calculate Energy Uncertainty

Given that the minimum uncertainty in the energy of the photons is 1.0\(\%\), calculate the uncertainty \( \Delta E \) as a percent of the photon energy. \[ \Delta E = 0.01 \times 3.18 \times 10^{-19} \text{ J} = 3.18 \times 10^{-21} \text{ J} \]

Step 4: Apply the Time-Energy Uncertainty Principle

Use the time-energy uncertainty principle: \[ \Delta E \Delta t \geq \frac{\hbar}{2} \] where \( \hbar = \frac{6.626 \times 10^{-34}}{2\pi} \) (reduced Planck's constant). Calculate \( \Delta t \) by rearranging the formula: \[ \Delta t \geq \frac{\hbar}{2\Delta E} \]

Step 5: Calculate the Minimum Pulse Duration

Using the values calculated: \[ \hbar = \frac{6.626 \times 10^{-34}}{2\pi} \approx 1.055 \times 10^{-34} \text{ J s} \] Substitute \( \Delta E = 3.18 \times 10^{-21} \text{ J} \) into the equation: \[ \Delta t \geq \frac{1.055 \times 10^{-34}}{2 \times 3.18 \times 10^{-21}} \] Calculating this gives \( \Delta t \approx 1.66 \times 10^{-14} \text{ s} \).

Key Concepts

Photon Energy

Photon energy is a fundamental concept in quantum physics that describes the energy carried by a single photon, a particle of light. Each photon has a specific amount of energy, which depends on its wavelength or frequency. The shorter the wavelength, the higher the energy of the photon. This relationship is given by the formula:\[ E = \frac{hc}{\lambda} \]Where:

• \( E \) is the energy of the photon.
• \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ J s})\).
• \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\).
• \( \lambda \) is the wavelength of the photon.

By substituting the values into this equation, one can find the energy of a photon, enabling the understanding of various phenomena like light emission and absorption, as explored in the exercise given.

Wavelength of Light

The wavelength of light is a key characteristic that defines the type and energy of the light. It is the distance between successive peaks of the light wave. Different wavelengths correspond to different colors in the visible light spectrum, with shorter wavelengths being closer to blue/violet and longer wavelengths near red. The relationship with energy is inversely proportional. This means that as the wavelength decreases, the energy of the photon increases, as shown in the formula:

• Short wavelength = High energy.
• Long wavelength = Low energy.

The exercise uses a wavelength of 625 nm (nanometers), which falls in the visible spectrum as an orange-red light. Understanding the wavelength also helps in determining the photon energy and other quantum mechanical properties.

Planck's Constant

Planck's constant is a fundamental physical constant denoted by \( h \). It is essential in quantum mechanics and embodies the quantization of energy. Planck's constant has the value of \( 6.626 \times 10^{-34} \text{ J s} \), and it plays a critical role in calculations involving photon energy and wave mechanics.\( h \) appears in the photon energy equation \( E = \frac{hc}{\lambda} \), thereby connecting the wave-like and particle-like nature of light. It represents the proportionality factor between the energy of a photon and the frequency of its associated electromagnetic wave. This constant forms the core of many quantum theories and is intrinsically linked to phenomena at atomic and subatomic levels.

Reduced Planck's Constant

The reduced Planck’s constant, denoted \( \hbar \) and pronounced as "h-bar," is derived from Planck’s constant \( h \) and is used in situations where angular frequency is involved. It is calculated by dividing Planck's constant by \( 2\pi \), thus:\[ \hbar = \frac{h}{2\pi} \approx 1.055 \times 10^{-34} \text{ J s} \]This constant simplifies expressions involving rotational systems and appears in equations involving quantizations, like the time-energy uncertainty principle:\[ \Delta E \Delta t \geq \frac{\hbar}{2} \]In this snippet from the exercise, \( \hbar \) facilitates the calculation of minimum pulse duration of light when considering the uncertainty in photon energy. It reflects the more nuanced behavior of particles at a quantum level, encapsulating the wave-particle duality of light.