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Calculate the de Broglie wavelength of (a) a \(50.0 \mathrm{~kg}\) woman jogging leisurely at \(2.0 \mathrm{~m} / \mathrm{s},\) (b) a free electron with kinetic energy \(2.0 \mathrm{MeV}\). and (c) a free electron with kinetic energy \(20 \mathrm{eV}\). Use the proper relativistic expression when necessary.

Short Answer

Expert verified
The de Broglie wavelength of (a) a \(50.0 \mathrm{~kg}\) woman jogging leisurely at \(2.0 \mathrm{~m/s}\) is very small and practically undetectable. For (b) an electron with kinetic energy \(2.0 \mathrm{~MeV}\) and (c) an electron with kinetic energy \(20 \mathrm{~eV}\), the wavelengths are larger and closer to what would be expected for a particle of such small mass.

Step by step solution

01

Calculate wavelength of a woman jogging

Firstly, calculate the De Broglie wavelength of the woman using the formula \(\lambda = \frac{h}{mv}\). Given, \(m = 50 kg\) and \(v = 2 m/s\). Insert these values into equation to find \(\lambda\). Use \(h = 6.626 \times 10^{-34} Js\) for Planck's constant.
02

Calculate wavelength of an electron with kinetic energy 2 MeV

Here, kinetic energy is given, \(E = 2 MeV\). However, this energy must be converted into joules. \(1eV = 1.6 \times 10^{-19}J\). Therefore, \(E = 2 \times 1.6 \times 10^{-13} J\). As this is more than 0.5 MeV, the relativistic energy formula should be used: \(E = \sqrt{(pc)^2 + (mc^2)^2}\). Solve for \(p\), then substitute into De Broglie's formula to get \(\lambda\).
03

Calculate wavelength of an electron with kinetic energy 20 eV

Repeat the steps as in Step 2, but here \(E = 20eV = 20 \times 1.6 \times 10^{-19} J\). Here the energy is less than \(0.5 MeV\), so the non-relativistic formula can be used: \(E = \frac{1}{2} mv^2\). Solve for \(v\) in this case, then substitute \(v\) into De Broglie's formula to find \(\lambda\).

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Key Concepts

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

Understanding Relativistic Energy
Relativistic energy is crucial when dealing with particles moving at speeds close to the speed of light. In such cases, classical physics fails to accurately describe the particle's behavior, and we must use Einstein's theory of relativity instead. The relativistic energy formula is given by:\[ E = \sqrt{(pc)^2 + (mc^2)^2} \]where:
  • \(E\) is the total energy of the particle,
  • \(p\) is the momentum,
  • \(c\) is the speed of light,
  • \(m\) is the rest mass of the particle.
This equation accounts for the increase in mass that occurs as a particle approaches the speed of light.
When calculating the de Broglie wavelength for particles like electrons, which can move at such high velocities, this formula helps us determine the momentum by solving for \(p\). Once we have \(p\), we can easily use de Broglie’s relation, \(\lambda = \frac{h}{p}\), to find the wavelength. This ensures accuracy when dealing with high-energy or fast-moving particles like the free electron with 2 MeV energy in our problem.
De Broglie Wavelength and Kinetic Energy
The kinetic energy of a particle is the energy it possesses due to its motion. For non-relativistic speeds, which are much slower than the speed of light, the kinetic energy \(E_k\) is given by:\[ E_k = \frac{1}{2}mv^2 \]where:
  • \(m\) is the mass,
  • \(v\) is the velocity.
In the context of de Broglie’s theory, knowing the kinetic energy allows us to find the velocity required for subsequent calculations of a particle's wavelength using the de Broglie relation.
For instance, when we consider an electron with a kinetic energy of 20 eV, its velocity can be calculated using the kinetic energy formula. This situation involves speeds where non-relativistic equations apply, simplifying the computations. Once the velocity is determined, it can be plugged into de Broglie’s equation, \(\lambda = \frac{h}{mv}\), to find the wavelength.
The Role of Planck's Constant
Planck's constant \(h\) is one of the pivotal constants in quantum mechanics. It is defined as \(6.626 \times 10^{-34} \mathrm{Js}\) and serves as the bridge between the wave and particle nature of matter. In de Broglie’s framework, Planck's constant helps connect a particle's momentum to its wavelength using the famous relation:\[ \lambda = \frac{h}{p} \]This relation implies that even massive objects have a wavelength, though it is typically extremely small and hard to detect for macroscopic objects, like the jogging woman in the problem.
Planck's constant ensures that we can calculate the de Broglie wavelength for any particle once we know its momentum. It emphasizes the duality nature, showing that matter has both particle and wave characteristics. This duality is especially pronounced and observable for small particles like electrons, which exhibit significant wavelike behavior due to their relatively small mass and large associated wavelengths. In the exercise, Planck's constant remains a crucial part of each wavelength calculation loop.

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Most popular questions from this chapter

(a) What accelerating potential is needed to produce electrons of wavelength \(5.00 \mathrm{nm} ?\) (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)?

Determine \(\lambda_{\mathrm{m}},\) the wavelength at the peak of the Planck distribution, and the corresponding frequency \(f,\) at these temperatures: (a) \(3.00 \mathrm{~K} ;\) (b) \(300 \mathrm{~K} ;\) (c) \(3000 \mathrm{~K}\).

A \(10.0 \mathrm{~g}\) marble is gently placed on a horizontal tabletop that is \(1,75 \mathrm{~m}\) wide. (a) What is the maximum uncertainty in the horizontal position of the marble? (b) According to the Heisenberg uncertainty principle, what is the minimum uncertainty in the horizontal velocity of the marble? (c) In light of your answer to part (b), what is the longest time the marble could remain on the table? Compare this time to the age of the universe, which is approximately 14 billion years. (Hint: Can you know that the horizontal velocity of the marble is exactly zero?)

(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength \(0.0600 \mathrm{nm} ?\) (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength \(0.0600 \mathrm{nm} ?\) (Hint: In each case the initial kinetic energy is negligible.)

CP CALC You have entered a contest in which the contestants drop a marble with mass \(20.0 \mathrm{~g}\) from the roof of a building onto a small target \(25.0 \mathrm{~m}\) below. From uncertainty considerations, what is the typical distance by which you'll miss the target, given that you aim with the highest possible precision? (Hint: The uncertainty \(\Delta x_{f}\) in the \(x\) -coordinate of the marble when it reaches the ground comes in part from the uncertainty \(\Delta x_{i}\) in the \(x\) -coordinate initially and in part from the initial uncertainty in \(v_{x}\). The latter gives rise to an uncertainty \(\Delta v_{x}\) in the horizontal motion of the marble as it falls. The values of \(\Delta x_{i}\) and \(\Delta v_{x}\) are related by the uncertainty principle. A small \(\Delta x_{i}\) gives rise to a large \(\Delta v_{x}\), and vice versa. Find the value of \(\Delta x_{i}\) that gives the smallest total uncertainty in \(x\) at the ground. Ignore any effects of air resistance.

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