A free proton has a wave function given by \(\Psi(x, t)=A e^{i\left(5.02 \times 10^{11} x-8.00 \times 10^{15} t\right)}\) The coefficient of \(x\) is inverse meters \(\left(\mathrm{m}^{-1}\right)\) and the coefficient on \(t\) is inverse seconds \(\left(\mathrm{s}^{-1}\right) .\) Find its momentum and energy.

Short Answer

Expert verified
The short answer is the momentum \(p = \dfrac{6.626 \times 10^{-34} \, \mathrm{J \cdot s}}{\dfrac{2\pi}{5.02 \times 10^{11} \, \mathrm{m}^{-1}}}\) and the energy \(E = (6.626 \times 10^{-34} \, \mathrm{J \cdot s}) \cdot \dfrac{8.00 \times 10^{15} \, \mathrm{s}^{-1}}{2\pi}\).

Step by step solution

01

Identify the wave function coefficients

The wave function is given by \[\Psi(x, t) = A e^{i(5.02 \times 10^{11} x - 8.00 \times 10^{15} t)}.\] The coefficient of \(x\) (in \(\mathrm{m}^{-1}\)) is \(5.02 \times 10^{11} \, \mathrm{m}^{-1}\), and the coefficient of \(t\) (in \(\mathrm{s}^{-1}\)) is \(8.00 \times 10^{15} \, \mathrm{s}^{-1}\).
02

Calculate the momentum

Using the de Broglie wavelength formula relating momentum and wavelength, we can extract the momentum information from the wave function. The formula is given by \(p = \dfrac{h}{\lambda}\), where \(p\) is the momentum, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \mathrm{J \cdot s}\)), and \(\lambda\) is the wavelength (in meters). Since the coefficient of \(x\) represents the wavenumber (\(k\)), which is equal to \(\dfrac{2\pi}{\lambda}\), we can first find the wavelength \(\lambda\): \[k = 5.02 \times 10^{11} \, \mathrm{m}^{-1} = \dfrac{2\pi}{\lambda}.\] Now, find out \(\lambda\): \[\lambda = \dfrac{2\pi}{5.02 \times 10^{11} \, \mathrm{m}^{-1}}.\] Finally, we can compute the momentum according to the de Broglie wavelength formula: \[p = \dfrac{h}{\lambda} = \dfrac{6.626 \times 10^{-34} \, \mathrm{J \cdot s}}{\lambda}.\]
03

Calculate the energy

To determine the energy, we can use the Planck-Einstein relation for photons: \(E = hf\), where \(f\) is the frequency of the wave (in seconds\(^{-1}\)) and \(E\) is the energy. The coefficient of \(t\) represents the angular frequency \(\omega\) (\(\mathrm{s}^{-1}\)), which is equal to \(2\pi f\). Thus, we can first find the frequency \(f\) : \[\omega = 8.00 \times 10^{15} \, \mathrm{s}^{-1} = 2\pi f.\] Now, find out \(f\): \[f = \dfrac{8.00 \times 10^{15} \, \mathrm{s}^{-1}}{2\pi}.\] Finally, we can compute the energy using the Planck-Einstein relation: \[E = hf = (6.626 \times 10^{-34} \, \mathrm{J \cdot s}) f.\] Now that we have derived the values for both the momentum and energy, we can plug in the numbers to find the answers. These will be the final values for the momentum and energy of the given free proton.

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Key Concepts

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

Wave Function
The wave function, often represented as \( \Psi(x, t) \), is a fundamental concept in quantum mechanics that encapsulates all the information about a quantum system's state. Imagining as the DNA of a particle, it provides a comprehensive description of the particle's behavior, such as its location and momentum probability distributions.

For example, a wave function like \( \Psi(x, t) = A e^{i(5.02 \times 10^{11} x - 8.00 \times 10^{15} t)} \) can seem mystifying at first glance. But let's decode it: \(|\Psi(x, t)|^2\), the square of the wave function's magnitude, tells us the probability density of finding the particle at a particular position and time. The expression \(|A|^2\) sees 'A' as the amplitude, which could be associated with the probability density when we square it.

In our exercise, the complex exponential component manifests the wave-like nature of quantum particles, also connecting the probabilities with the nicely predictable patterns of waves. This ties back to the fact that particles at the quantum level behave like both particles and waves, an idea at the heart of wave-particle duality.
De Broglie Wavelength
The de Broglie wavelength, denoted as \( \lambda \), is another revolutionary concept in quantum mechanics introduced by Louis de Broglie. This idea is a shining example of wave-particle duality. It proposes that particles can exhibit wave-like properties and that each particle has a wavelength inversely proportional to its momentum.

Using the formula \( p = \frac{h}{\lambda} \), where 'p' is the momentum and 'h' is Planck's constant, we can determine a particle's momentum given its wavelength. In the case of our exercise, the wavenumber 'k', taken from the coefficient of 'x' in the wave function, is related to the de Broglie wavelength by \( k = \frac{2\pi}{\lambda} \). Therefore, once we determine \( \lambda \), we can calculate the particle's momentum, a bridge between wave-like and particle-like properties, showcasing the versatility of quantum mechanics in describing the microcosmic world.
Planck-Einstein Relation
The Planck-Einstein relation is a pivotal equation in quantum mechanics linking the energy of photons with their frequency, using the equation \( E = hf \), where 'E' is energy, 'h' is Planck's constant, and 'f' is the frequency of the wave.

In the realm of quantum mechanics, where quantization is king, this formula implies that energy comes in discrete packets, called quanta. This discrete nature demarks a clear departure from classical mechanics where energy is seen as a continuous flow. For our proton exercise, the angular frequency \( \omega \) is directly related to the frequency by \( \omega = 2\pi f \) and is thus critical for calculating the energy of the particle.

Moreover, the Planck-Einstein relation embodies the quantization of electromagnetic waves, a key principle that underlies the understanding of phenomena such as the photoelectric effect and Planck's black body radiation law. It's through applying this relation that we can unfold the quantum world's intricacies, like determining the energy of a particle represented in wave function, and appreciate the quantized harmony of the universe's most fundamental levels.

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Most popular questions from this chapter

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