/*! 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} Q65PE The velocity of a proton in an a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Q65PE (page 1064)

The velocity of a proton in an accelerator is known to an accuracy of \[{\bf{0}}{\bf{.250 \% }}\]of the speed of light. (This could be small compared with its velocity.) What is the smallest possible uncertainty in its position?

Short Answer

Expert verified

The electron鈥檚 minimum uncertainty in velocity is\[4.2 \times {10^{ - 14}}\;{\rm{m}}\]

Step by step solution

01

Determine the formulas:

Determine the uncertainty principle formula:

\(\Delta x\Delta p = \frac{h}{{2\pi }}\)

Here,\(\Delta x\)is the uncertainty in the position and\(\Delta p\)is the uncertainty in the momentum.

Consider the expression for the momentum is\[\Delta p = m\Delta v\].

02

Evaluate the electron’s minimum uncertainty in velocity

Determine the value of uncertainty in position:

\[\begin{array}{c}\Delta x = \frac{h}{{4\pi m\Delta v}}\\ = \frac{{6.63 \times {{10}^{ - 34}}{\rm{ J}} \cdot {\rm{s}}}}{{4\pi \times 1.67 \times {{10}^{ - 27}}{\rm{ kg}} \times 2.5 \times {{10}^{ - 3}} \times 3.0 \times {{10}^8}\;{\rm{m}}{{\rm{s}}^{ - 1}}}}\\ = 4.2 \times {10^{ - 14}}\;{\rm{m}}\end{array}\]

Therefore, the velocity is:\[4.2 \times {10^{ - 14}}\;{\rm{m}}\].

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 91影视!

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

Question: Violet light of wavelength \[{\rm{400 nm}}\] ejects electrons with a maximum kinetic energy of \[{\rm{0}}{\rm{.860 eV}}\] from sodium metal. What is the binding energy of electrons to sodium metal?

A certain heat lamp emits 200 W of mostly IR radiation averaging 1500 nm in wavelength.

(a) What is the average photon energy in joules?

(b) How many of these photons are required to increase the temperature of a person's shoulder by 2.0oC, assuming the affected mass is 4.0 kg with a specific heat of 0.83 kcal/kgoC.

{\vphantom {{\;{\bf{kcal}}} {{\bf{kg}}^\circ {\bf{C}}}}} \right.

\kern-\nulldelimiterspace} {{\bf{kg}}^\circ {\bf{C}}}}\]. Also assume no other significant heat transfer.

(c) How long does this take?


Which aspects of the photoelectric effect cannot be explained without photons? Which can be explained without photons? Are the latter inconsistent with the existence of photons?

How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of \({\rm{1}}{\rm{.00 - W}}\)? Assume the average energy per photon is \({\rm{75}}{\rm{.0 - keV}}\).

(a) Calculate the number of photoelectrons per second ejected from a \(1.00\,{\rm{m}}{{\rm{m}}^{\rm{2}}}\) area of sodium metal by \(500\,{\rm{nm EM}}\) radiation having an intensity of \(1.30\,{\rm{kW/}}{{\rm{m}}^{\rm{2}}}\) (the intensity of sunlight above the Earth鈥檚 atmosphere). (b) Given that the binding energy is\(2.28\,{\rm{eV}}\), what power is carried away by the electrons? (c) The electrons carry away less power than brought in by the photons. Where does the other power go? How can it be recovered?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Electricity

Read Explanation

Astrophysics

Read Explanation

Thermodynamics

Read Explanation

Physics of Motion

Read Explanation

Measurements

Read Explanation
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.