Explanation 1) Concept: Necessary condition for circular motion can be used to find the speed of electron moving in magnetic field because we know the product of field and radius of path. Also we know the expression of maximum kinetic energy of electron, which can be used to find work required to remove the electron from the gold atom. 2) Formula: i) Radius of circle traced by electron in presence magnetic field r = m e v e B ii) Kinetic energy of electron K = 1 2 m e v 2 iii) Work required to remove the electron from the atom W = ϵ - K m a x iv) Energy of photon ϵ = h c λ 3) Given: i) Wavelength of x ray, λ = 71 p m = 71 × 10 - 3 n m ii) B r = 1.88 × 10 - 4 T m where r is the radius of circular path. 4) Calculations: a) The electron will move in a circular orbit if the magnetic force provides the necessary centripetal force. Therefore, e B v = m e v 2 r where r , m e , a n d v are radius, mass and speed of electron in the circular orbit respectively

10th Edition

ISBN: 9781118230725

Author: David Halliday, Robert Resnick, Jearl Walker

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Concept explainers

Particle Theory of Light

The particle theory of light was the proposal made by Newton in 1704 in his treatise Opticks. This is the most basic light theory, in which light is thought to be made up of microscopic particles known as "corpuscles" and that's why this particle theory …

Stopping Potential

In an experiment conducted by Heinrich Hertz, an apparatus was made where the incident light was made to fall on the metallic plate, it was discovered that metals emit electrons. The surface electrons are bound to metals with a minimum amount of energy a…

Quantization of Charges

An electron is a negatively charged subatomic particle either attached to an an atom or sticks to the nucleus of the atom. Electrons exert the negative charge that tries to balance the positive charge of the nucleus.

Question

Chapter 38, Problem 21P

To determine

To find:

a) Maximum kinetic energy of electron

b) Work done in removing electron from the gold atom

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 38, Problem 20P

Chapter 38, Problem 22P

Students have asked these similar questions

What is the velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic fields. E⃗ E→ and B⃗ B→ are also perpendicular to each other and have magnitudes 4900 V/m and 6.7×10−3 T , respectively. What is the radius of the electron orbit if the electric field is turned off?

X rays with a wavelength of 73.0 pm are directed onto a gold foil and eject tightly bound electrons from atoms of the foil. The ejected electrons then move in circular paths of radius r in a region of uniform magnetic field with magnitude B. For the fastest of the ejected electrons, the product Br is equal to 2.46 x 104 T•m. Find (a) the maximum kinetic energy of those electrons and (b) the work done in removing them from the gold atoms.

An alpha particle with velocity v = (3 x 105,0,0) m/s enters a region where themagnetic field has a value B = (0,0,1.2) T. Determine the required magnitude and directionof an electric field E that will allow the alpha particle to continue to move along the x axis.

Chapter 38 Solutions

Fundamentals of Physics Extended

Ch. 38 - Prob. 1Q Ch. 38 - Prob. 2Q Ch. 38 - Prob. 3Q Ch. 38 - Prob. 4Q Ch. 38 - Prob. 5Q Ch. 38 - Prob. 6Q Ch. 38 - Prob. 7Q Ch. 38 - Prob. 8Q Ch. 38 - Prob. 9Q Ch. 38 - Prob. 10Q

Ch. 38 - Prob. 11Q Ch. 38 - Prob. 12Q Ch. 38 - Prob. 13Q Ch. 38 - Prob. 14Q Ch. 38 - Prob. 15Q Ch. 38 - Prob. 16Q Ch. 38 - Prob. 1P Ch. 38 - Prob. 2P Ch. 38 - Prob. 3P Ch. 38 - Prob. 4P Ch. 38 - Prob. 5P Ch. 38 - Prob. 6P Ch. 38 - Prob. 7P Ch. 38 - Prob. 8P Ch. 38 - Prob. 9P Ch. 38 - Prob. 10P Ch. 38 - Prob. 11P Ch. 38 - Prob. 12P Ch. 38 - Prob. 13P Ch. 38 - Prob. 14P Ch. 38 - Prob. 15P Ch. 38 - Prob. 16P Ch. 38 - Prob. 17P Ch. 38 - Prob. 18P Ch. 38 - Prob. 19P Ch. 38 - Prob. 20P Ch. 38 - Prob. 21P Ch. 38 - Prob. 22P Ch. 38 - Prob. 23P Ch. 38 - Prob. 24P Ch. 38 - Prob. 25P Ch. 38 - Prob. 26P Ch. 38 - Prob. 27P Ch. 38 - Prob. 28P Ch. 38 - Prob. 29P Ch. 38 - Prob. 30P Ch. 38 - Prob. 31P Ch. 38 - Prob. 32P Ch. 38 - Prob. 33P Ch. 38 - Prob. 34P Ch. 38 - Prob. 35P Ch. 38 - Prob. 36P Ch. 38 - Prob. 37P Ch. 38 - Prob. 38P Ch. 38 - Prob. 39P Ch. 38 - Prob. 40P Ch. 38 - Prob. 41P Ch. 38 - Prob. 42P Ch. 38 - Prob. 43P Ch. 38 - Prob. 44P Ch. 38 - Prob. 45P Ch. 38 - Prob. 46P Ch. 38 - Prob. 47P Ch. 38 - Prob. 48P Ch. 38 - Prob. 49P Ch. 38 - Prob. 50P Ch. 38 - Prob. 51P Ch. 38 - Prob. 52P Ch. 38 - Prob. 53P Ch. 38 - Prob. 54P Ch. 38 - Prob. 55P Ch. 38 - Prob. 56P Ch. 38 - Prob. 57P Ch. 38 - Prob. 58P Ch. 38 - Prob. 59P Ch. 38 - Prob. 60P Ch. 38 - Prob. 61P Ch. 38 - Prob. 62P Ch. 38 - Prob. 63P Ch. 38 - Prob. 64P Ch. 38 - Prob. 65P Ch. 38 - Prob. 66P Ch. 38 - Prob. 67P Ch. 38 - Prob. 68P Ch. 38 - Prob. 69P Ch. 38 - Prob. 70P Ch. 38 - Prob. 71P Ch. 38 - Prob. 72P Ch. 38 - Prob. 73P Ch. 38 - Prob. 74P Ch. 38 - Prob. 75P Ch. 38 - Prob. 76P Ch. 38 - Prob. 77P Ch. 38 - Prob. 78P Ch. 38 - Prob. 79P Ch. 38 - Prob. 80P Ch. 38 - Prob. 81P Ch. 38 - Prob. 82P Ch. 38 - Prob. 83P Ch. 38 - Prob. 84P Ch. 38 - Prob. 85P Ch. 38 - Prob. 86P Ch. 38 - Prob. 87P Ch. 38 - Prob. 88P Ch. 38 - Prob. 89P Ch. 38 - Prob. 90P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

a) Maximum kinetic energy of electron b) Work done in removing electron from the gold atom

Solution Summary: The author explains how the required condition for circular motion can be used to find the speed of electron moving in magnetic field.

Concept explainers

To find: a) Maximum kinetic energy of electron b) Work done in removing electron from the gold atom

Want to see the full answer?

Chapter 38 Solutions

To find:

a) Maximum kinetic energy of electron

b) Work done in removing electron from the gold atom