Problem 47 - College Physics

Question

\(\bullet\) A certain atom has an energy level 3.50 eV above the ground state. When excited to this state, it remains \(4.0 \mu s,\) on the average, before emitting a photon and returning to the ground state. (a) What is the energy of the photon? What is its wavelength? (b) What is the smallest possible uncertainty in energy of the photon?

Step-by-Step Solution

Answer

(a) 3.50 eV, 354 nm; (b) 8.25x10^-11 eV.

Step 1: Energy of the Photon

The energy of the photon emitted is equal to the energy difference between the excited state and the ground state. Given that the energy level in the excited state is 3.50 eV above the ground state, the energy of the photon is 3.50 eV.

Step 2: Convert Energy to Joules

To calculate the wavelength of the photon, first convert the energy from electron volts to joules. Use the conversion factor: 1 eV = \(1.602 \times 10^{-19}\) J.\[E = 3.50 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 5.607 \times 10^{-19} \, \text{J}\]

Step 3: Calculate the Wavelength

Use the formula that relates photon energy to wavelength:\[\lambda = \frac{hc}{E}\]where \(h\) is Planck's constant \(6.626 \times 10^{-34} \, \text{Js}\) and \(c\) is the speed of light \(3.00 \times 10^8 \, \text{m/s}\).Calculate \(\lambda\):\[\lambda = \frac{6.626 \times 10^{-34} \, \text{Js}\times 3.00 \times 10^8 \, \text{m/s}}{5.607 \times 10^{-19} \, \text{J}} \approx 3.54 \times 10^{-7} \, \text{m} = 354 \, \text{nm}\]

Step 4: Calculate Uncertainty in Energy

According to the energy-time uncertainty principle:\[\Delta E \cdot \Delta t \geq \frac{\hbar}{2}\]where \(\hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34} \, \text{Js}\). Given \(\Delta t = 4.0 \, \mu s = 4.0 \times 10^{-6} \, \text{s}\), the smallest possible uncertainty in energy is:\[\Delta E \geq \frac{1.055 \times 10^{-34} \, \text{Js}}{2 \times 4.0 \times 10^{-6} \, \text{s}} \approx 1.32 \times 10^{-29} \, \text{J}\]Convert this back to eV:\[\Delta E \geq \frac{1.32 \times 10^{-29} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 8.25 \times 10^{-11} \, \text{eV}\]

Key Concepts

Energy Levels

In atomic physics, energy levels refer to the fixed energies that electrons within an atom can possess. These are usually visualized as steps in a stairway, with the ground state being the lowest step. Electrons reside in the ground state under normal conditions. When they absorb energy, they can "jump" to a higher energy level or excited state. However, these excited states are unstable, so electrons tend to "fall" back to the ground state eventually, releasing energy in the form of photons. This released energy is equal to the difference between the two energy levels. In our example, an atom in the 3.50 eV excited state will release a photon with 3.50 eV of energy when it returns to the ground state.

Photon Wavelength

The wavelength of a photon is inversely related to its energy. Higher energy photons have shorter wavelengths. To find a photon's wavelength, use the formula: \[\lambda = \frac{hc}{E}\]Here, \(h\) is Planck's constant \(6.626 \times 10^{-34} \, \text{Js}\), and \(c\) is the speed of light \(3.00 \times 10^8 \, \text{m/s}\). The example shows a photon's energy of 3.50 eV, which first converts to joules.

• Convert the energy: \(3.50 \, \text{eV} = 5.607 \times 10^{-19} \, \text{J}\)

You can then determine that the photon's wavelength is approximately 354 nm. Shorter wavelengths fall in the visible spectrum or beyond, such as ultraviolet, depending on the energy levels involved.

Energy Uncertainty

Heisenberg's uncertainty principle tells us that certain pairs of properties, like energy and time, cannot both be known to an arbitrary precision. This means there's an inherent uncertainty involved when determining the exact energy of a system in a given time frame. In mathematical terms: \[\Delta E \cdot \Delta t \geq \frac{\hbar}{2}\]Here, \(\Delta E\) is the uncertainty in energy, \(\Delta t\) is the uncertainty or duration of time, and \(\hbar\) is the reduced Planck's constant \(\frac{h}{2\pi}\). In our exercise, the excited state lasts an average of \(4.0 \, \mu\text{s}\), which provides an uncertainty in energy of at least \(8.25 \times 10^{-11} \, \text{eV}\). The smaller the time interval \(\Delta t\), the higher the uncertainty in energy, and vice versa.

Planck's Constant

Planck's constant is a fundamental figure in quantum mechanics, signifying the proportionality between the energy of a photon and the frequency of its electromagnetic wave. Represented as \(h\), it has the constant value \(6.626 \times 10^{-34} \, \text{Js}\). This constant plays a pivotal role in deriving various scientific formulas such as the energy of a photon \(E\) given by \(E = hf\), where \(f\) is frequency.

• Relates energy to frequency: \(E = hf\)
• Used in wavelength relation: \(\lambda = \frac{hc}{E}\)
• Integral to the uncertainty principle

Planck's constant helps bridge macroscopic classical physics and microscopic quantum physics, showing how discrete quantities of energy undergo transitions.