Question

A certain atom has an energy state 3.50 eV above the ground state. When excited to this state, the atom remains for 2.0 \(\mu\)s, on average, before it emits a photon and returns to the ground state. (a) What are the energy and wavelength of the photon? (b) What is the smallest possible uncertainty in energy of the photon?

Short answer

(a) Energy = 3.50 eV, Wavelength = 355 nm. (b) Uncertainty in energy = \(5.26 \times 10^{-29}\) J.

Step-by-step solution

Understanding the Energy of the Photon

Since the atom emits a photon to return to the ground state, the energy of the photon corresponds to the energy difference between these states. This is given as 3.50 eV.

Converting Energy Units

Convert the energy from electronvolts (eV) to joules (J) using the conversion factor: \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\). Thus, \(3.50 \text{ eV} = 3.50 \times 1.602 \times 10^{-19} \text{ J} = 5.607 \times 10^{-19} \text{ J}\).

Calculating the Wavelength of the Photon

Use the formula relating energy and wavelength: \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34} \text{ J·s}\) is Planck's constant and \(c = 3.00 \times 10^8 \text{ m/s}\) is the speed of light. Solving for \(\lambda\), we have:\[\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{5.607 \times 10^{-19}} \approx 3.55 \times 10^{-7} \text{ m} \text{ or } 355 \text{ nm}\]

Understanding Uncertainty in Energy

According to the energy-time uncertainty principle, \(\Delta E \cdot \Delta t \geq \frac{\hbar}{2}\), where \(\hbar = \frac{h}{2\pi}\). In this case, \(\Delta t = 2.0 \times 10^{-6} \text{ s}\).

Calculating the Uncertainty in Energy

Calculate \(\Delta E\) using \(\Delta E \geq \frac{h}{4\pi \Delta t}\). Substituting the values gives:\[\Delta E \geq \frac{6.626 \times 10^{-34}}{4\pi \times 2.0 \times 10^{-6}} \approx 5.26 \times 10^{-29} \text{ J} \]

Key concepts

Energy Levels

Energy levels in an atom represent the specific amounts of energy an electron can have. These levels define the possible states or orbits that electrons can occupy around a nucleus. Each energy level corresponds to a specific energy value, measured in electronvolts (eV) or joules (J).
When an electron moves between these levels, it must absorb or emit energy equal to the difference between these levels. In this exercise, the atom has an energy state that is 3.50 eV above the ground state. This means that for an electron to reach this excited state from the ground level, it needs to absorb energy equivalent to 3.50 eV.
Understanding energy levels is crucial in atomic physics because they dictate how atoms absorb and release energy, influencing chemical reactions and emission spectra.

Photon Emission

Photon emission occurs when an electron transitions from a higher energy level to a lower one, emitting energy as a photon in the process. The energy of this photon matches the energy difference between the initial and final states.
For the problem we're examining, the atom emits a photon as it transitions to the ground state from an excited state. This emitted photon carries 3.50 eV, the energy it has to release to return to the ground level.
To find the wavelength of an emitted photon, we can use the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the photon energy, \( h \) is Planck's constant, and \( c \) is the speed of light. This relationship underscores the connection between a photon's energy and its wavelength, which is essential in understanding light and electromagnetic radiation.

Uncertainty Principle

The uncertainty principle is a fundamental concept in quantum mechanics, famously introduced by Werner Heisenberg. It states that certain pairs of physical properties, like position and momentum, cannot both be precisely measured simultaneously. In the context of this problem, it refers to energy and time.
The energy-time uncertainty principle is given by \( \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \), where \( \Delta E \) is the uncertainty in energy and \( \Delta t \) is the uncertainty in time. Here, \( \hbar \) is the reduced Planck's constant, equal to \( \frac{h}{2\pi} \).
For the excited atom, remaining on average for 2.0 \( \mu \)s before emitting a photon, \( \Delta t \) represents the average lifetime of the excited state. This principle is pivotal in quantum physics, indicating that there's a limit to the precision with which certain physical quantities can be known.

Electronvolt to Joules Conversion

In atomic physics, energy is often measured in electronvolts (eV). An electronvolt is a unit of energy equal to approximately \( 1.602 \times 10^{-19} \) joules. It is often used in the field because it is a convenient scale for processes involving electrons in atoms.
In the given exercise, the energy difference is presented as 3.50 eV. To convert it to joules for further calculations, we use the following conversion: \( 3.50 \text{ eV} = 3.50 \times 1.602 \times 10^{-19} \text{ J} \), resulting in \( 5.607 \times 10^{-19} \text{ J} \).
This conversion is essential for expressing energy in the International System of Units (SI), making calculations consistent and comparable with other scientific measurements.