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In atomic physics, energy levels refer to the different energies that an electron in an atom can have. Each level represents a specific energy state that an electron may occupy, typically structured in a series from the ground state up to higher levels. The ground state is the lowest energy state, and higher energy states are termed excited states.

In the given example, the atom in question has an excited energy level that is 3.50 electron volts (eV) above the ground state. When an electron moves from this excited state back to the lower ground state, it emits a photon with an energy equal to the difference between these two states.

• This emitted photon corresponds to the precise energy transition from one level to another.
• Knowing the exact energy levels helps us understand the emission or absorption spectrum of an atom.

This concept is crucial in fields like spectroscopy, where the energy levels of elements determine the unique spectrum of light they emit or absorb.

The wavelength of a photon relates directly to its energy and can be calculated using Planck's equation: \[\lambda = \frac{hc}{E}\]where:

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

To find the wavelength, we first need to convert the energy from electron volts to joules, since energy in eV can be expressed as:\[E = 3.50 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 5.607 \times 10^{-19} \text{ J}\]Substitute this value into the wavelength formula to find:\[\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5.607 \times 10^{-19}} = 3.56 \times 10^{-7} \text{ m}\]Thus, the wavelength of the emitted photon is approximately 356 nm, which places it within the visible range of the electromagnetic spectrum.

The uncertainty principle, formulated by Werner Heisenberg, is a fundamental concept in quantum mechanics. It states that certain pairs of physical properties, like position and momentum, cannot be simultaneously measured to arbitrarily high precision. In this exercise, we apply it to energy \((\Delta E)\) and time \((\Delta t)\): \[\Delta E \Delta t \geq \frac{\hbar}{2}\]where:

• \(\hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34} \text{ Js}\)
• \(\Delta t = 4.0 \times 10^{-6} \text{ s}\)

In this context, the principle implies that as the photon remains in the excited state for an average time \(\Delta t\), there is an inherent uncertainty in the exact energy of the photon \(\Delta E\):\[\Delta E \geq \frac{1.055 \times 10^{-34}}{2 \times 4.0 \times 10^{-6}} = 1.318 \times 10^{-29} \text{ J}\]Converted back to electron volts, the uncertainty is \(8.23 \times 10^{-11} \text{ eV}\). This small uncertainty illustrates limitations in measuring energy due to the brief time the electron is in the excited state.