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A proton has a kinetic energy of \(1.0 \mathrm{MeV}\). If its momentum is measured with an uncertainty of \(5.0 \%\), what is the minimum uncertainty in its position?

Short Answer

Expert verified
The minimum uncertainty in the proton's position is calculated via the Heisenberg Uncertainty Principle, resulting in \(\Delta x\).

Step by step solution

01

Calculate momentum from kinetic energy

The kinetic energy (K) is given as 1.0 MeV, which can be converted to Joules (J) by multiplying by a factor of \(1.6 \times 10^{-13}\). The mass of the proton (m) is \(1.67 \times 10^{-27}\) Kg. Equation \(K = p^2 / (2m)\) can be solved for the momentum, \(p = \sqrt{2Km}\).
02

Calculate the uncertainty in momentum

The uncertainty in the measurement of momentum is given by the problem as 5.0%. This uncertainty in momentum (\(\Delta p\)) can be found by multiplying the momentum by 0.05.
03

Calculate the minimum uncertainty in position

Finally, solve the Heisenberg Uncertainty Principle for the uncertainty in position. The solution for \(\Delta x\) is \(\Delta x \geq \hbar / (2 \Delta p)\), where \(\hbar\) is the reduced Planck constant, approximately \(1.055 \times 10^{-34}\) Js.

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Key Concepts

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

Kinetic Energy of Particles
Kinetic energy represents the energy that a particle has due to its motion. It is a fundamental concept in both classical and quantum physics and can be expressed using the formula \( K = \frac{1}{2}mv^2 \), where \( K \) is the kinetic energy, \( m \) is the mass, and \( v \) is the velocity of the particle.

In the context of quantum mechanics and the Heisenberg Uncertainty Principle, the kinetic energy of particles like protons can be used to derive other physical properties such as momentum. Because a proton's kinetic energy and mass are known, the momentum \( p \) is calculated using the relation \( p = \sqrt{2Km} \), where \( K \) is the kinetic energy converted to Joules and \( m \) is the mass of the proton. This step is pivotal in understanding how measurements at the quantum level, such as the kinetic energy, influence other properties like position and momentum.
Proton Momentum
Momentum is a measure of the quantity of motion of a moving body and in classical physics is given by the product of its mass and velocity. However, in the quantum world, it relates to the wave-like nature of particles and is subject to Heisenberg's Uncertainty Principle.

In the exercise, we are concerned with the proton's momentum \( p \), which is derived from its kinetic energy. When the momentum is measured with an uncertainty, it implies that the exact value is not precisely known. The uncertainty of the proton's momentum \( \Delta p \) can be quantified as a percentage of its calculated momentum. By multiplying the proton's momentum by the uncertainty percentage; we get the value of \( \Delta p \). This uncertainty in momentum is crucial for determining the corresponding uncertainty in the proton's position, a direct application of the Heisenberg Uncertainty Principle.
Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. A fundamental element of quantum mechanics is Heisenberg's Uncertainty Principle, which states that certain pairs of physical properties, like position and momentum, cannot both be known to arbitrary precision. The more precisely one property is measured, the less precisely the other can be controlled or predicted.

This principle has profound implications for the nature of the quantum world, as it implies a fundamental limit to what can be known about a system's physical properties. The principle is mathematically expressed as \( \Delta x \Delta p \geq \frac{\hbar}{2} \), where \( \Delta x \) is the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( \hbar \) is the reduced Planck's constant.

In the exercise, after calculating the uncertainty in momentum, we apply this principle to find the minimum uncertainty in position. This illustrates an interesting aspect of quantum mechanics: the act of measuring affects the system, limiting the exactness of our knowledge about a particle's properties.

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Most popular questions from this chapter

Calculate the de Broglie wavelength of a proton that is accelerated through a potential difference of \(10 \mathrm{MV}\).

A woman on a ladder drops small pellets toward a spot on the floor. (a) Show that, according to the uncertainty principle, the miss distance must be at least $$ \Delta x=\left(\frac{\hbar}{2 m}\right)^{1 / 2}\left(\frac{H}{2 g}\right)^{1 / 4} $$ where \(H\) is the initial height of each pellet above the floor and \(m\) is the mass of each pellet. (b) If \(H=2.0 \mathrm{~m}\) and \(m=0.50 \mathrm{~g}\), what is \(\Delta x ?\)

A matter wave packet. (a) Find and sketch the real part of the matter wave pulse shape \(f(x)\) for a Gaussian amplitude distribution \(a(k)\), where $$ a(k)=A e^{-\alpha^{2}\left(k-k_{0}\right)^{2}} $$ Note that \(a(k)\) is peaked at \(k_{0}\) and has a width that decreases with increasing \(\alpha .\) (Hint: In order to put \(f(x)=(2 \pi)^{-1 / 2} \int_{-\infty}^{+\infty} a(k) e^{i k x} d k \quad\) into the standard form \(\int_{-\infty}^{+\infty} e^{-a z^{2}} d z\), complete the square in \(\left.k_{.}\right)\) (b) \(\mathrm{By}\) comparing the result for the real part of \(f(x)\) to the standard form of a Gaussian function with width \(\Delta x\), \(f(x) \propto A e^{-(x / 2 \Delta x)^{2}}\), show that the width of the matter wave pulse is \(\Delta x=\alpha \cdot(c)\) Find the width \(\Delta k\) of \(a(k)\) by writing \(a(k)\) in standard Gaussian form and show that \(\Delta x \Delta k=\frac{1}{2}\), independent of \(\alpha .\)

Robert Hofstadter won the 1961 Nobel prize in physics for his pioneering work in scattering 20 -GeV electrons from nuclei. (a) What is the \(\gamma\) factor for a \(20-\mathrm{GeV}\) electron, where \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2} ?\) What is the momentum of the electron in \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) ? (b) What is the wavelength of a \(20-\mathrm{GeV}\) electron and how does it compare with the size of a nucleus?

An air rifle is used to shoot \(1.0-\mathrm{g}\) particles at \(100 \mathrm{~m} / \mathrm{s}\) through a hole of diameter \(2.0 \mathrm{~mm}\). How far from the rifle must an observer be to see the beam spread by \(1.0 \mathrm{~cm}\) because of the uncertainty principle? Compare this answer with the diameter of the Universe \(\left(2 \times 10^{26} \mathrm{~m}\right)\)

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