Problem 4 - Physics for Scientists and Engineers with Modern Physics

Question

(1) An electron remains in an excited state of an atom for typically \(10^{-8} \mathrm{s}\) . What is the minimum uncertainty in the energy of the state (in \(\mathrm{eVV}\) )

Step-by-Step Solution

Answer

The minimum uncertainty in energy is approximately \(3.29 \times 10^{-8}\) eV.

Step 1: Identify the Relevant Formula

The minimum uncertainty in energy can be estimated using Heisenberg's Uncertainty Principle for energy and time, which states that \( \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \). Here, \( \Delta E \) is the uncertainty in energy, \( \Delta t \) is the uncertainty in time, and \( \hbar \) is the reduced Planck's constant approximately equal to \( 1.0545718 \times 10^{-34} \) Js.

Step 2: Substitute Known Values into the Formula

Substitute \( \Delta t = 10^{-8} \) seconds and \( \hbar = 1.0545718 \times 10^{-34} \) Joule-seconds into the formula:\[ \Delta E \cdot 10^{-8} \geq \frac{1.0545718 \times 10^{-34}}{2} \].

Step 3: Solve for \( \Delta E \)

Rearrange the inequality to solve for \( \Delta E \):\[ \Delta E \geq \frac{1.0545718 \times 10^{-34}}{2 \times 10^{-8}} \].Calculate \( \Delta E \):\[ \Delta E \geq 5.272859 \times 10^{-27} \text{ Joules} \].

Step 4: Convert Energy to Electronvolts

Convert the uncertainty in energy from Joules to electronvolts (eV) using the conversion factor \( 1 \text{ Joule} = 6.242 \times 10^{18} \text{ eV} \):\[ \Delta E \geq 5.272859 \times 10^{-27} \times 6.242 \times 10^{18} \text{ eV} \].Calculate the result:\[ \Delta E \geq 3.29 \times 10^{-8} \text{ eV} \].

Key Concepts

Energy Uncertainty

Heisenberg's Uncertainty Principle plays a crucial role in the quantum mechanics field. It provides the framework to understand the limitations of measuring certain pairs of variables simultaneously, like position and momentum, or energy and time.
To get to grips with the energy uncertainty, consider the principle stating that there is always some inherent minimum uncertainty in energy and time measurements, depicted by the formula:\[ \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \]

• \(\Delta E\) represents the uncertainty in energy.
• \(\Delta t\) symbolizes the uncertainty in time.
• \(\hbar\) is the reduced Planck's constant, approximately equal to \(1.0545718 \times 10^{-34}\) J·s.

In practical terms, it suggests that if an electron remains in an excited state for a time \(\Delta t = 10^{-8} \, \text{s}\), then the uncertainty in its energy—our \(\Delta E\)—is dictated by the equation. Replacing the time uncertainty and solving for the energy uncertainty is what the exercise focuses on.Understanding this principle underscores that in the quantum realm, perfect accuracy in measurements is inherently impossible.

Excited State Lifetime

In quantum physics, the term "excited state" refers to an energy state of an atom that is higher than its ground state. When an electron is excited, it absorbs energy and moves to a higher energy level.
The lifetime of these excited states is often very short, and in the given exercise, it is approximately \(10^{-8} \, \text{s}\). This small time window indicates how quick an electron may return to its ground state and is crucial for determining the uncertainty in energy due to its direct relationship with \(\Delta t\).

• Longer lifetimes imply smaller energy uncertainty.
• Shorter lifetimes result in greater energy uncertainty.

Linking this to Heisenberg's principle, when the electron quickly drops back to its ground state, the extent of energy uncertainty increases, because there's less time to make precise measurements. Understanding the excited state's duration provides perspective on how transient these energy levels are and how they fit into broader quantum behaviors.

Electronvolt Conversion

Electronvolts (eV) are a unit of energy commonly used in atomic and particle physics. Converting measurements into electronvolts helps contextualize energy levels in these fields in a more workable scale.
The problem involves converting the calculated energy uncertainty from Joules to electronvolts. The conversion factor is given by:

• \(1 \text{ Joule} = 6.242 \times 10^{18} \text{ eV}\).

Multiplying the minimum energy uncertainty \(\Delta E\) by this factor transforms the value into a more usable unit for practical applications in physics. For instance, industries like semiconductor technology or cosmic ray analysis find electronvolt measurements extremely convenient.
This conversion not only aids in simplifying complex scientific problems but also provides a consistent and clear unit of measurement for energies at the quantum level, thereby easing communication and interpretation within scientific communities.