/*! 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 41 The speed of an electron having ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 41

The speed of an electron having a wavelength of the order of \(1 \AA\) will be (A) \(7.25 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (B) \(6.26 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (C) \(5.25 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (D) \(4.24 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The speed of an electron having a wavelength of the order of \(1 \AA\) will be approximately \(7.25 \times 10^6 \, \text{m/s}\) (Option A).

Step by step solution

01

Write down the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \(\lambda = \dfrac{h}{p}\) Where \(\lambda\) is the wavelength, \(h\) is the Planck constant (\(6.626 \times 10^{-34} \, \text{Js}\)), and \(p\) is the particle's momentum.
02

Write down the relationship between momentum and velocity

The momentum of a particle is given by the product of its mass and velocity: \(p = m_e v\) Where \(m_e\) is the mass of the electron (\(9.109 \times 10^{-31} \, \text{kg}\)) and \(v\) is its velocity.
03

Combine the two formulas to relate wavelength and velocity

We'll now eliminate momentum from the equations by substituting the equation for momentum into the de Broglie wavelength formula: \(\lambda = \dfrac{h}{m_e v}\)
04

Solve for velocity

To find the velocity of the electron, we'll rearrange the equation to solve for \(v\): \(v = \dfrac{h}{m_e \lambda}\)
05

Plug in the given values and compute the velocity

Substituting the given wavelength (\(\lambda = 1 \, \text{脜} = 1 \times 10^{-10} \, \text{m}\)), Planck constant (\(h = 6.626 \times 10^{-34} \, \text{Js}\)), and electron mass (\(m_e = 9.109 \times 10^{-31} \, \text{kg}\)), we get: \(v = \dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1 \times 10^{-10}}\) \(v \approx 7.27 \times 10^6 \,\text{m/s}\) Comparing this value to the provided options, we find that the closest answer is: (A) \(7.25 \times 10^6 \, \text{m/s}\)

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

The speed of daughter nuclei is (A) \(c \sqrt{\frac{\Delta m}{M+\Delta m}}\) (B) \(c \frac{\Delta m}{M+\Delta m}\) (C) \(c \sqrt{\frac{2 \Delta m}{M}}\) (D) \(c \sqrt{\frac{\Delta m}{M}}\)

Electromagnetic radiation whose electric component varies with time as \(E=C_{1}\left(C_{2}+C_{3} \cos \omega t\right) \cos \omega_{0} t\), here \(C_{1}, C_{2}\) and \(C_{3}\) are constants, is incident on lithium and liberates photoelectrons. Let the kinetic energy of most energetic electrons be \(0.592 \times 10^{-19} \mathrm{~J}\). Given that \(\omega_{0}=3.6 \times 10^{15} \mathrm{rad} / \mathrm{s}\) and \(\omega=6 \times 10^{14}\) \(\mathrm{rad} / \mathrm{sec} .\) The work function of lithium is (take planks constant \(\left.h=6.6 \times 10^{-34} \mathrm{MKS}\right)\). (A) \(1.2 \mathrm{eV}\) (B) \(1.5 \mathrm{eV}\) (C) \(2.1 \mathrm{eV}\) (D) \(2.39 \mathrm{eV}\)

If \(N_{0}\) is the original mass of the substance of half-life period \(t_{1 / 2}=5\) years, then the amount of substance left after 15 years is (A) \(N_{0} / 8\) (B) \(N_{0} / 16\) (C) \(N_{0} / 2\) (D) \(N_{0} / 4\)

Assertion: Work function of aluminium is \(4.2 \mathrm{eV}\). If two photons of each of energy \(2.5 \mathrm{eV}\) strike on an electron of aluminium, the electron is not emitted. Reason: In photoelectric effect a single photon interacts with a single electron and electron is emitted only if energy of each of incident photon is greater than the work function. (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\overline{\mathrm{C}}\) (D) D

Which of the following atoms has the lowest ionization potential? (A) \({ }_{7}^{14} N\) (B) \({ }_{55}^{133} \mathrm{Cs}\) (C) \({ }_{18}^{40} A r\) (D) \({ }_{8}^{16} O\)
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Waves Physics

Read Explanation

Electricity

Read Explanation

Torque and Rotational Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Engineering Physics

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.