General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
bartleby

Concept explainers

Question
Book Icon
Chapter 27, Problem 1RQ
To determine

The idea for composition of atom using Rutherford’s experiment.

Expert Solution & Answer
Check Mark

Answer to Problem 1RQ

Rutherford experiment led to an idea that the nucleus of the atom is composed of massive positively charged particles surrounded by electrons.

Explanation of Solution

Rutherford experiment shows that maximum alpha particles are scattered with small angle of deviation, some scattering in large angles, and very few are reflected back towards the source from the target when energetic positively charged alpha particles is thrown towards the thin gold foil.

Rutherford explains this incident in a way that the small volume of the target atom is occupied by the positively charged nucleus and the negatively charged electrons occupies rest of the volume of the target.

This led to an idea that the nucleus of the atom is composed of massive positively charged particles surrounded by electrons.

Conclusion:

Therefore, Rutherford experiment led to an idea that the nucleus of the atom is composed of massive positively charged particles surrounded by electrons.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Someone wanted to build a scale model of the atom with a nucleus 1.4 m in diameter.  How far away would the nearest electron need to be in meters? Assume the orbital radius of an electron is 10^(-10)m, while the radius of the nucleus is 10^(-15) m.  ra=?
Suppose someone wanted to build a scale model of the atom with a nucleus 1.4 m in diameter.  How far away would the nearest electron need to be in meters? Assume the orbital radius of an electron is 10^(-10)m, while the radius of the nucleus is 10^15 m.  ra=?
Use the below values for this problem. Please note that the mass for H is for the entire atom (proton & electron). Neutron: m,= 1.67493x1027 kg= 1.008665 u = 939.57 MeVIC H: my = 1.67353x10 27 kg = 1.007825 u = 938.78 MeVic 1u= 1.6605x10-27 kg = 931.5 MeVic? Consider the following decay: 211 At 207 Bi + a. 211 At has a mass of 210.9874963 u, 207 Bi has a mass of 206.981593 u, and a has a mass of 4.002603 u. 85 83 85 83 Determine the disintegration energy (Q-value) in MeV. Determine the binding energy (in MeV) for 211 At. 85 EB =
Knowledge Booster
Background pattern image
Physics
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.
Similar questions
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Text book image
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
Text book image
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
Text book image
Intro Spectroscopy
Physics
ISBN:9781305221796
Author:PAVIA
Publisher:Cengage
Text book image
An Introduction to Physical Science
Physics
ISBN:9781305079137
Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:Cengage Learning