Problem 13 - Physics for Scientists and Engineers with Modern Physics

Question

(II) An electron in the \(n=2\) state of hydrogen remains there on average about \(10^{-8}\) s before jumping to the \(n=1\) state. (a) Estimate the uncertainty in the energy of the \(n=2\) state. (b) What fraction of the transition energy is this? (c) What is the wavelength, and width (in \(\mathrm{nm}\) ), of this line in the spectrum of hydrogen?

Step-by-Step Solution

Answer

(a) \( \Delta E \approx 5.27 \times 10^{-27} \text{ J} \); (b) Fraction \( \approx 3.23 \times 10^{-9} \); (c) Wavelength \( \approx 121.6 \text{ nm} \), Width \( \approx 1.25 \times 10^{-5} \text{ nm} \).

Step 1: Applying the Energy-Time Uncertainty Principle

To find the uncertainty in the energy of the state, we use the energy-time uncertainty principle: \[ \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \]Given \(\Delta t = 10^{-8}\, \text{s}\), we use \(\hbar = 1.0545718 \times 10^{-34} \, \text{J}\cdot\text{s}\):\[ \Delta E \approx \frac{1.0545718 \times 10^{-34}}{2 \times 10^{-8}} \approx 5.2729 \times 10^{-27} \, \text{J} \].

Step 2: Calculating Transition Energy

Calculate the energy difference between the \(n=2\) and \(n=1\) states using the formula:\[ E_n = - \frac{13.6}{n^2} \text{ eV} \]The transition energy \(\Delta E_{12}\) is thus:\[ \Delta E_{12} = \frac{13.6}{1} - \frac{13.6}{4} = 10.2 \, \text{eV} = 1.632 \times 10^{-18} \, \text{J}\].

Step 3: Fraction of Uncertainty in Transition Energy

Calculate the fraction of the uncertainty in the energy relative to the transition energy:\[ \text{Fraction} = \frac{\Delta E}{\Delta E_{12}}= \frac{5.2729 \times 10^{-27}}{1.632 \times 10^{-18}} \approx 3.23 \times 10^{-9}\].

Step 4: Calculating Wavelength of Transition

Use the relation between energy and wavelength:\[ \Delta E_{12} = \frac{hc}{\lambda} \]Solve for \(\lambda\):\[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{1.632 \times 10^{-18}} \approx 1.216 \times 10^{-7} \, \text{m} = 121.6 \, \text{nm}\].

Step 5: Calculating Line Width in Spectrum

The uncertainty in energy \(\Delta E\) leads to a spectral line width \(\Delta \lambda\) via:\[ \Delta \lambda = \frac{\lambda^2 \cdot \Delta E}{hc} \]\[ \Delta \lambda \approx \frac{(1.216 \times 10^{-7})^2 \times 5.2729 \times 10^{-27}}{6.626 \times 10^{-34} \times 3 \times 10^{8}} \approx 1.25 \times 10^{-5} \, \text{nm}\].

Key Concepts

Energy-Time Uncertainty Principle

The Energy-Time Uncertainty Principle is one of the fascinating aspects of quantum physics, which speaks about the limit of precision when measuring energy and time. According to the principle, there is a fundamental limit to how precisely one can measure both the energy and the duration of a quantum state. This is expressed in the formula: \[ \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \] where \(\Delta E\) is the uncertainty in energy, \(\Delta t\) is the uncertainty in time, and \(\hbar\) is the reduced Planck's constant, approximately \(1.0545718 \times 10^{-34}\, \text{J}\cdot\text{s}\). In practice, this principle implies that short-lived states, like excited states of an electron, have a large intrinsic energy uncertainty. For an electron in the \(n=2\) state of hydrogen, which remains there for about \(10^{-8}\) seconds, we can calculate the energy uncertainty as: \[ \Delta E \approx \frac{1.0545718 \times 10^{-34}}{2 \times 10^{-8}} \approx 5.2729 \times 10^{-27} \, \text{J} \] This value tells us about the natural spread of possible energies the electron state could have.

Quantum Transitions

Quantum transitions are movements of electrons between different energy levels within an atom. These transitions become observable through the emission or absorption of light at specific wavelengths, which corresponds to the energy difference between the initial and final states. For a hydrogen atom transitioning from the \(n=2\) state to the \(n=1\) state, the energy difference can be calculated using the formula: \[ E_n = - \frac{13.6}{n^2} \text{ eV} \] Here, converting the energy from electron volts to joules, the energy difference \(\Delta E_{12}\) becomes: \[ \Delta E_{12} = \frac{13.6}{1} - \frac{13.6}{4} = 10.2 \, \text{eV} = 1.632 \times 10^{-18} \, \text{J} \] The fascinating part about quantum transitions is their precision; every time the same transition occurs, the exact same amount of energy is released.

Wavelength Calculation

The relationship between energy and wavelength is at the heart of understanding spectral lines. According to the formula: \[ \Delta E_{12} = \frac{hc}{\lambda} \] where \(h\) is Planck's constant \(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\), \(c\) is the speed of light \(3 \times 10^{8} \, \text{m/s}\), and \(\lambda\) is the wavelength. Solving for \(\lambda\), we can find the wavelength of the spectral line associated with the transition: \[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{1.632 \times 10^{-18}} \approx 1.216 \times 10^{-7} \, \text{m} = 121.6 \, \text{nm} \] In spectroscopy, the width of spectral lines can also be informative. It is influenced by the energy uncertainty from the transition. Using the approximation, the spectral line width \(\Delta \lambda\) can be derived as: \[ \Delta \lambda \approx \frac{(1.216 \times 10^{-7})^2 \times 5.2729 \times 10^{-27}}{6.626 \times 10^{-34} \times 3 \times 10^{8}} \approx 1.25 \times 10^{-5} \, \text{nm} \] This very slight broadening of the spectral line provides critical insights into precise measurements and quantum behaviors.