Problem 6 - Physics for Scientists and Engineers with Modern Physics

Question

(1) The lifetime of a typical excited state in an atom is about 10 ns. Suppose an atom falls from one such excited state and emits a photon of wavelength about 500 \(\mathrm{nm}\) . Find the fractional energy uncertainty \(\Delta E / E\) and wavelength uncertainty \(\Delta \lambda / \lambda\) of this photon.

Step-by-Step Solution

Answer

Fractional energy uncertainty \( \Delta E / E \) is approximately \( 1.33 \times 10^{-8} \) and wavelength uncertainty \( \Delta \lambda / \lambda \) is approximately \( 1.33 \times 10^{-8} \).

Step 1: Understanding the Problem

We are asked to find the fractional energy uncertainty \( \Delta E / E \) and the wavelength uncertainty \( \Delta \lambda / \lambda \) for a photon emitted by an atom transitioning from an excited state.

Step 2: Identify Given Data

The lifetime of the excited state is \( \Delta t = 10 \text{ ns} = 10 \times 10^{-9} \text{ s} \) and the wavelength of the photon emitted is \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} \).

Step 3: Use Uncertainty Principle for Energy

The energy-time uncertainty principle is given by \( \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \), where \( \hbar = \frac{h}{2\pi} \approx 1.0545718 \times 10^{-34} \text{ J}\cdot\text{s} \).

Step 4: Calculate \(\Delta E\)

Using \( \Delta t = 10^{-8} \text{ s} \), we find \( \Delta E \geq \frac{1.0545718 \times 10^{-34}}{2 \times 10^{-8}} \), which simplifies to \( \Delta E \geq 5.3 \times 10^{-27} \text{ J} \).

Step 5: Calculate Energy \(E\) of Photon

Energy \(E\) is given by \( E = \frac{hc}{\lambda} \), where \( h = 6.62607015 \times 10^{-34} \text{ J}\cdot\text{s} \) and \( c = 3 \times 10^8 \text{ m/s} \). Substituting \( \lambda = 500 \times 10^{-9} \text{ m} \), we calculate \( E \approx 3.972 \times 10^{-19} \text{ J} \).

Step 6: Calculate \(\frac{\Delta E}{E}\)

Use the relation \( \frac{\Delta E}{E} = \frac{5.3 \times 10^{-27}}{3.972 \times 10^{-19}} \approx 1.33 \times 10^{-8} \).

Step 7: Use Energy-Wavelength Relation

The wavelength uncertainty \( \Delta \lambda \) is derived from \( \Delta E = \frac{hc}{\lambda^2} \cdot \Delta \lambda \).

Step 8: Calculate \(\Delta \lambda \)

Rearrange \( \Delta E = \frac{hc}{\lambda^2} \cdot \Delta \lambda \) to \( \Delta \lambda = \frac{\Delta E \cdot \lambda^2}{hc} \) and substitute \( \Delta E \) and \( \lambda \).

Step 9: Compute \(\Delta \lambda / \lambda \)

Using \( \Delta \lambda = 6.63 \times 10^{-12} \text{ m} \), find \( \frac{\Delta \lambda}{\lambda} = \frac{6.63 \times 10^{-12}}{500 \times 10^{-9}} \approx 1.326 \times 10^{-8} \).

Key Concepts

Energy-Time Uncertainty

At the heart of quantum mechanics lies the intriguing principle of uncertainty. Among its many manifestations, the energy-time uncertainty is one of the key concepts that governs atomic transitions. This principle can be expressed as \( \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \). It tells us that the product of the uncertainty in energy (\( \Delta E \)) and the uncertainty in time (\( \Delta t \)) must be greater than or equal to half of the reduced Planck constant \( \hbar \).

This principle comes into play when atoms transition from an excited state back to a lower energy state. The lifetime of an excited state (\( \Delta t \)) plays a crucial role in determining the energy's uncertainty. For instance, a shorter-lived excited state would imply a larger \( \Delta E \). In the exercise, we are given \( \Delta t = 10 \text{ ns} \), leading to the calculation of \( \Delta E \geq 5.3 \times 10^{-27} \text{ J} \).

This uncertainty not only highlights the fundamental limits of precision in measuring energy over time but also dictates the precision with which we can ascertain other properties of emitted particles, like their wavelength.

Photon Emission

When an atom transitions from a higher energy state to a lower one, it emits a photon. This process is central to the study of light and atomic interactions. The energy of the photon is given by \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the emitted photon.

Understanding photon emission helps us predict how atoms release energy and the kind of light they produce. In this exercise, we investigate photons with a wavelength of \( 500 \text{ nm} \). Using the known constants, we calculated the energy of this emitted photon to be approximately \( 3.972 \times 10^{-19} \text{ J} \).

One important aspect of photon emission is its link to the energy-time uncertainty. The uncertainty in the energy of the photon (\( \Delta E \)) directly affects how precisely we can measure the wavelength, making the understanding of this process essential for predicting light behavior in various scientific fields.

Wavelength Uncertainty

Wavelength, a core characteristic of light, influences how we perceive and utilize light in practical applications. According to the energy-wavelength relation, the uncertainty in energy (\( \Delta E \)) and wavelength (\( \Delta \lambda \)) are interconnected. This relation is expressed by \( \Delta E = \frac{hc}{\lambda^2} \cdot \Delta \lambda \).

The exercise guides us to compute \( \Delta \lambda \,/\, \lambda \), representing the fractional wavelength uncertainty. It's calculated from the earlier determined \( \Delta E \) and the given \( \lambda \) value. This calculation results in \( \Delta \lambda \approx 6.63 \times 10^{-12} \text{ m} \), and \( \Delta \lambda / \lambda \approx 1.326 \times 10^{-8} \).

This means that while the wavelength can be precisely measured, there's always an inherent uncertainty due to the quantum mechanical principles. Comprehending wavelength uncertainty is crucial for applications requiring high precision, such as spectroscopy and quantum computing, where slight changes in wavelength can lead to significant variations in outcomes.