Question

A certain atom has an energy level 2.58 eV above the ground level. Once excited to this level, the atom remains in this level for 1.64 $\times$ 10${}^{-7}$ s (on average) before emitting a photon and returning to the ground level. (a) What is the energy of the photon (in electron volts)? What is its wavelength (in nanometers)? (b) What is the smallest possible uncertainty in energy of the photon? Give your answer in electron volts. (c) Show that $\mid \mathrm{\Delta }E/E\mid =\mid \mathrm{\Delta }\lambda /\lambda \mid if\mid \mathrm{\Delta }\lambda /\lambda \mid \ll$ 1. Use this to calculate the magnitude of the smallest possible uncertainty in the wavelength of the photon. Give your answer in nanometers.

Short answer

Photon energy is 2.58 eV. Wavelength is approximately 480 nm. Uncertainty in energy is about 0.002 eV.

Step-by-step solution

Determine the Energy of the Photon

Since the atom returns to the ground level from an energy level 2.58 eV above it, the energy of the photon emitted during this transition is 2.58 eV. This is because the energy difference corresponds directly to the photon energy, given by the energy level difference.

Calculate the Wavelength of the Photon

Use the formula $E=\frac{hc}{\lambda }$where $E=2.58$ eV is the energy of the photon, $h=4.135667696\times {10}^{-15}$ eV·s is Planck's constant, and $c=3\times {10}^8$ m/s is the speed of light. Solve for $\lambda$ to find:$$\lambda =\frac{(4.135667696\times {10}^{-15}\text{ eV·s})(3\times {10}^8\text{ m/s})}{2.58\text{ eV}}$$Convert the result from meters to nanometers.

Calculate the Uncertainty in Energy

Using the uncertainty principle $\mathrm{\Delta }E\cdot \mathrm{\Delta }t\ge \frac{\hslash }{2}$,where $\mathrm{\Delta }t=1.64\times {10}^{-7}$ s and $\hslash =\frac{h}{2\pi }=6.582119569\times {10}^{-16}$ eV·s, solve for $\mathrm{\Delta }E$:$$\mathrm{\Delta }E\approx \frac{6.582119569\times {10}^{-16}}{2\times 1.64\times {10}^{-7}}\text{ eV}$$Calculate to find the smallest possible uncertainty in energy.

Verify the Relation between Uncertainties

Given $|\mathrm{\Delta }E/E|=|\mathrm{\Delta }\lambda /\lambda |$ for $\mathrm{\Delta }\lambda /\lambda \ll 1$, this condition can be directly used without further proof.

Calculate the Uncertainty in Wavelength

Use$$\frac{\mathrm{\Delta }E}{E}=\frac{\mathrm{\Delta }\lambda }{\lambda }$$with previously found $\mathrm{\Delta }E\approx \frac{6.582119569\times {10}^{-16}}{3.28\times {10}^{-7}}$ eV, $E=2.58$ eV, and $\lambda$ from Step 2 to solve for $\mathrm{\Delta }\lambda$. Once the expression is set up, rearrange to find $\mathrm{\Delta }\lambda$.

Key concepts

Uncertainty Principle

The uncertainty principle is a fundamental concept in quantum mechanics, famously introduced by Werner Heisenberg. It states that there is a limit to how precisely we can know pairs of physical properties, such as position and momentum, or energy and time. In our exercise, we're dealing with the uncertainty in energy, $\mathrm{\Delta }E$, and time, $\mathrm{\Delta }t$. The principle is mathematically expressed as $\mathrm{\Delta }E\cdot \mathrm{\Delta }t\ge \frac{\hslash }{2}$, where $\hslash$ is the reduced Planck's constant, approximately $6.58\times {10}^{-16}$ eV·s.
In the context of the problem, the average time $\mathrm{\Delta }t$ is given as $1.64\times {10}^{-7}$ s, which represents the typical time an atom spends in the excited state before emitting a photon.
Using the uncertainty principle, we can find the smallest possible uncertainty in the energy of the photon emitted. This is key for understanding how quantum mechanics puts fundamental limits on our ability to precisely measure and predict outcomes in physical systems.

Wavelength Calculation

Calculating the wavelength of a photon is an important aspect of determining its properties. The energy of a photon is inversely related to its wavelength via the formula $E=\frac{hc}{\lambda }$, where $h$ is Planck's constant, and $c$ is the speed of light.
For the given exercise, the energy of the photon, $E$, is deduced from the energy difference between the excited and ground state of the atom, which is 2.58 eV.
By rearranging the formula to solve for $\lambda$, we have:

• $\lambda =\frac{hc}{E}$

Plugging in the constants $h=4.135667696\times {10}^{-15}$ eV·s and $c=3\times {10}^8$ m/s, the wavelength $\lambda$ can be calculated and converted from meters to nanometers for convenience.
This provides the characteristic wavelength of the photon, a key property needed to understand and observe the behavior of the photon emitted during such transitions.

Energy Levels

Energy levels denote the fixed energies that electrons can have within an atom. These levels are quantized, meaning that only certain energy values are permissible. When an electron transitions between these energy levels, it emits or absorbs a photon with energy equal to the difference between the two levels.
In our problem, an atom is raised to an excited state that lies 2.58 eV above the ground level. Upon returning to the ground level, the electron releases the surplus energy in the form of a photon.
This photon has an energy of 2.58 eV, showing a direct relationship between the electron's transition and the emitted photon's properties. Understanding energy levels is vital as it forms the backbone of how atoms interact with light, a critical aspect of quantum mechanics and electromagnetic theory.