/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Calculate the de Broglie wavelen... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(\lambda\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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)?

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

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature \(T\) and a diameter 3.0 times that of the hotter star. (a) What is the temperature of the hotter star in terms of \(T ?\) (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

\(\mathrm{CP}\) (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 \(\left.K=E-m c^{2} .\right)\) (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 beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops \(6.50 \times 10^{-14} \mathrm{~m}\) away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is \(6.64 \times 10^{-27} \mathrm{~kg} .\) (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in \(\mathrm{MeV}\). (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.