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An atom in a metastable state has a lifetime of 5.2 \(\mathrm{ms}\) . What is the uncertainty in energy of the metastable state?

Short Answer

Expert verified
The uncertainty in energy is approximately \( 1.0148 \times 10^{-32} \, \text{J} \).

Step by step solution

01

Understand the Uncertainty Principle

The uncertainty in energy and the lifetime of a state are related by the energy-time uncertainty principle, which is given by the equation \( \Delta E \Delta t \geq \frac{\hbar}{2} \), where \( \Delta E \) is the uncertainty in energy, \( \Delta t \) is the lifetime, and \( \hbar \) is the reduced Planck's constant.
02

Identify Known Values

We know from the problem that the lifetime \( \Delta t \) of the metastable state is \( 5.2 \, \text{ms} \) (which is \( 5.2 \times 10^{-3} \, \text{s} \)). The reduced Planck's constant \( \hbar \) is approximately \( 1.0545718 \times 10^{-34} \, \text{J} \cdot \text{s} \).
03

Rearrange the Equation

To find the uncertainty in energy, \( \Delta E \), we rearrange the uncertainty principle formula to \( \Delta E \geq \frac{\hbar}{2 \Delta t} \). This allows us to directly calculate \( \Delta E \) using the known values.
04

Calculate Uncertainty in Energy

Substitute the known values into the rearranged formula: \( \Delta E = \frac{1.0545718 \times 10^{-34} \, \text{J} \cdot \text{s}}{2 \times 5.2 \times 10^{-3} \, \text{s}} \). This will give us the uncertainty in energy, \( \Delta E \), in units of Joules.
05

Perform the Calculation

Compute the expression: \( \Delta E = \frac{1.0545718 \times 10^{-34}}{2 \times 5.2 \times 10^{-3}} \approx 1.0148 \times 10^{-32} \, \text{J} \).

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Key Concepts

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

Metastable State
In physics, a metastable state is a temporary energy state of an atom or a molecule that is only relatively stable. It can exist for a longer duration than an ordinary excited state but is not as stable as the ground state. This state allows the atom or molecule to exist in a higher energy level before eventually transitioning to a lower energy level or the ground state by releasing energy.
Metastable states are important in various physical phenomena, such as lasing in lasers and phosphorescence in materials. The unique property of these states is that they can last significantly longer, sometimes up to milliseconds or more, which is crucial for processes that depend on time-delayed reactions.
When considering applications and the importance of metastable states, remember that:
  • They allow energy to be stored temporarily in an atom or molecule.
  • These states enable processes such as phosphorescence, where light emission occurs long after the initial excitation.
  • They are pivotal in the operation of lasers, where energy is stored until triggered to release coherently.
Uncertainty in Energy
The concept of "Uncertainty in Energy" arises from the famous Heisenberg's Uncertainty Principle, specifically the energy-time uncertainty relation. This principle suggests that there is a fundamental limit to how precisely we can predict the energy of a system over a specific time period.
In the context of quantum mechanics, the uncertainty in energy \( \Delta E \) represents the range of potential energies that a metastable state can possess. Since the energy cannot be precisely determined, all the possible energy values fall within a range defined by this uncertainty.
The energy-time uncertainty principle is given by:\[\Delta E \Delta t \geq \frac{\hbar}{2}\]Where:
  • \( \Delta E \): Uncertainty in energy
  • \( \Delta t \): Lifetime of the state
  • \( \hbar \): Reduced Planck's constant, approximately \( 1.0545718 \times 10^{-34} \, \text{J} \cdot \text{s} \)
To find the uncertainty in energy, you rearrange the equation to:\[\Delta E \geq \frac{\hbar}{2 \Delta t}\]This rearrangement helps us calculate \( \Delta E \) when \( \Delta t \) and \( \hbar \) are known. For instance, if the lifetime \( \Delta t \) is given as \( 5.2 \, \text{ms} \), you can estimate the uncertainty in energy as shown in the solution steps by substituting the known values.
Planck's Constant
Planck's constant is a fundamental constant in physics that plays a crucial role in quantum mechanics. It is denoted by \( h \) and is essential in describing the sizes of quanta in quantum physics. The reduced Planck's constant, symbolized as \( \hbar \), is often used in formulas, especially when discussing angular momentum or action in quantum mechanics.
Planck's constant facilitates the quantization of energy levels for particles at the atomic and subatomic level. It is through this quantization that distinct energy levels, such as those seen in atoms and molecules, are possible.
Key points about Planck's constant:
  • It is approximately \( 6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \) for \( h \), and \( \hbar \) is \( 1.0545718 \times 10^{-34} \, \text{J} \cdot \text{s} \).
  • Planck's constant is central to Heisenberg's Uncertainty Principle, enabling us to describe inherent uncertainties in the properties like position and energy.
  • It establishes a link between energy and frequency, given by the equation \( E = h u \), where \( u \) is the frequency of the radiation.
Understanding Planck's constant is integral to grasping how energy levels work in quantum mechanics, and why certain states such as metastable states exist.

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Most popular questions from this chapter

(a)A particle with mass \(m\) has kinetic energy equal to three times its rest energy. What is the de Broglie wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: \(E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}\) and \(K=\) \(E-m c^{2} \cdot )\) (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

(a) What is the smallest amount of energy in electron volts that must be given to a hydrogen atom initially in its ground level so that it can emit the \(H_{\alpha}\) line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the \(n=3\) level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

Light from an ideal spherical blackbody 15.0 \(\mathrm{cm}\) in diameter is analyzed using a diffraction grating having 3850 lines/cm. When you shine this light through the grating, you observe that the peak-intensity wavelength forms a first-order bright fringe at \(\pm 11.6^{\circ}\) from the central bright fringe. (a) What is the temperature of the blackbody? (b) How long will it take this sphere to radiate 12.0 \(\mathrm{MJ}\) of energy?

What is the de Broglie wavelength for an electron with speed (a) \(v=0.480 c\) and \(v=0.960 c ?\) (Hint: Use the correct relativistic expression for linear momentum if necessary.)

Bohr Orbits of a Satellite. A 20.0 -kg satellite circles the earth once every 2.00 \(\mathrm{h}\) in an orbit having a radius of 8060 \(\mathrm{km} .\) (a) Assuming that Bohr's angular-momentum result \((L=n h / 2 \pi)\) applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number \(n\) of the orbit of the satellite. (b) Show from Bohr's angular momentum result and Newton's law of gravitation that the radius of an earth-satellite orbit is directly proportional to the square of the quantum number, \(r=k n^{2},\) where \(k\) is the constant of proportionality. (c) Using the result from part \((\mathrm{b}),\) find the distance between the orbit of the satellite in this problem and its next "allowed" orbit. (Calculate a numerical value.) (d) Comment on the possibility of observing the separation of the two adjacent orbits. (e) Do quantized and classical orbits correspond for this satellite? Which is the "correct" method for calculating the orbits?

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