Chapter 39: Problem 78
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
Step by step solution
Understanding the Energy of the Photon
Converting Energy Units
Calculating the Wavelength of the Photon
Understanding Uncertainty in Energy
Calculating the Uncertainty in Energy
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Energy Levels
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
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 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 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.