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An atom in a metastable state has a lifetime of 5.2 ms. What is the uncertainty in energy of the metastable state?

Short Answer

Expert verified
The energy uncertainty is approximately \(1.0148 \times 10^{-32} \) joules.

Step by step solution

01

Understanding Lifetime and Energy Uncertainty

The energy uncertainty of a quantum state, such as a metastable state of an atom, can be determined using the Heisenberg uncertainty principle. This principle in relation to energy (\(\Delta E\)) and time (\(\Delta t\)) is given as: \(\Delta E \cdot \Delta t \geq \frac{\hbar}{2}\) where \(\hbar\) is the reduced Planck’s constant (\(\hbar = 1.0545718 \times 10^{-34} J\cdot s\)). Here, \(\Delta t\) is provided as the lifetime of the metastable state.
02

Set Up the Inequality for Uncertainty

Given the relation \(\Delta E \cdot \Delta t = \frac{\hbar}{2}\), we can solve for the uncertainty in energy \(\Delta E\). \(\Delta t = 5.2 \) ms = \(5.2 \times 10^{-3} \) s.
03

Solve for Energy Uncertainty

To find \(\Delta E\), rearrange the formula: \(\Delta E = \frac{\hbar}{2 \cdot \Delta t}\). Substituting the values, we get: \[\Delta E = \frac{1.0545718 \times 10^{-34}}{2 \times 5.2 \times 10^{-3}}\].
04

Calculating the Result

Perform the calculation: \[\Delta E \approx \frac{1.0545718 \times 10^{-34}}{2 \times 5.2 \times 10^{-3}} \approx 1.0148 \times 10^{-32}\ J\]. Therefore, the uncertainty in energy of the metastable state is approximately \(1.0148 \times 10^{-32} \) joules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Metastable State
A metastable state refers to a temporary state of an atom where it is in an excited configuration. Unlike typical excited states, a metastable state has an unusually long lifetime before decaying to a more stable state. This prolonged existence makes it interesting to scientists as it provides a window to study various phenomena in quantum mechanics, such as energy levels and transitions.

The extended lifetime of a metastable state is due to restricted transition possibilities to a lower energy state. Such restrictions often occur when transitions to lower energy levels are "forbidden" or highly improbable under normal circumstances. This phenomenon is a key reason why some atomic states can last milliseconds, like the one mentioned with a 5.2 ms lifetime, in contrast to typical excited states that decay in nanoseconds.

Metastable states are incredibly significant in areas like lasers and phosphorescence, where sustained emission of light or energy over time is harnessed for practical applications.
Energy Uncertainty
Energy uncertainty is a principle that arises from the Heisenberg uncertainty principle, which posits that it is impossible to know both the energy and duration of a quantum system's state with absolute precision. In the context of a metastable state, this means that the exact energy level of the state cannot be perfectly defined if the lifetime of the state is known, and vice versa.

This uncertainty is expressed through the formula \[ \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \] where \( \Delta E \) is the energy uncertainty, \( \Delta t \) is the time uncertainty or the lifetime of the state, and \( \hbar \) is the reduced Planck's constant. The inequality highlights how knowledge about either time or energy impacts the precision with which the other can be determined.

Understanding energy uncertainty is crucial in quantum mechanics as it underscores the intrinsic limitations in measuring quantum states, influencing theories that tackle the behavior of atoms and particles at the smallest scales.
Reduced Planck's Constant
The reduced Planck's constant, often denoted by \( \hbar \), is a fundamental physical constant crucial in quantum mechanics. It is defined as the Planck's constant \( h \) divided by \( 2\pi \), or \( \hbar = \frac{h}{2\pi} \approx 1.0545718 \times 10^{-34} \, J \cdot s \).

\( \hbar \) is widely used as it naturally appears in quantum physics equations that describe the properties and behaviors of particles at quantum levels. In the context of the uncertainty principle, \( \hbar \) bridges the connection between the measurable energy of a system and the time duration over which it can be accurately measured.

This constant is pivotal in understanding the scale at which quantum effects become significant. It plays a critical role in formulas that govern quantum entities like electrons in an atom or photons in light, signifying its ubiquitous presence in the equations that describe the microscopic universe.

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Most popular questions from this chapter

(a) An electron moves with a speed of 4.70 \(\times\) 10\(^6\) m/s. What is its de Broglie wavelength? (b) A proton moves with the same speed. Determine its de Broglie wavelength.

(a) What accelerating potential is needed to produce electrons of wavelength 5.00 nm? (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

Suppose that the uncertainty of position of an electron is equal to the radius of the \(n\) = 1 Bohr orbit for hydrogen. Calculate the simultaneous minimum uncertainty of the corresponding momentum component, and compare this with the magnitude of the momentum of the electron in the \(n\) = 1 Bohr orbit. Discuss your results.

(a) If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each. (b) If a photon and an electron each have the same wavelength of 250 nm, find the energy of each. (c) You want to study an organic molecule that is about 250 nm long using either a photon or an electron microscope. Approximately what wavelength should you use, and which probe, the electron or the photon, is likely to damage the molecule the least?

A large number of neon atoms are in thermal equilibrium. What is the ratio of the number of atoms in a 5\(s\) state to the number in a 3\(p\) state at (a) 300 K; (b) 600 K; (c) 1200 K? The energies of these states, relative to the ground state, are E\(_{5s}\) = 20.66 eV and E\(_{3p}\) = 18.70 eV. (d) At any of these temperatures, the rate at which a neon gas will spontaneously emit 632.8-nm radiation is quite low. Explain why.

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