Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 149

In the beginning of the twentieth century, some scientists thought that a nucleus may contain both electrons and protons. Use the Heisenberg uncertainty principle to show that an electron cannot be confined within a nucleus. Repeat the calculation for a proton. Comment on your results. Assume the radius of a nucleus to be \(1.0 \times 10^{-15} \mathrm{~m} .\) The masses of an electron and a proton are \(9.109 \times 10^{-31} \mathrm{~kg}\) and \(1.673 \times 10^{-27} \mathrm{~kg},\) respectively. (Hint: Treat the diameter of the nucleus as the uncertainty in position.)

Short Answer

Expert verified
The calculation using the Heisenberg Uncertainty Principle shows that the minimum speed of an electron in the nucleus exceeds the speed of light, which is physically impossible. In contrast, a proton within the nucleus would need to move at a much smaller speed, making it possible. Thus, it validates the modern understanding that a nucleus contains protons, not electrons.

Step by step solution

01

Analyze given data

Identify the given values: the diameter of a nucleus (\(2.0 \times 10^{-15}\) m), the masses of an electron and proton (\(9.109 \times 10^{-31}\) kg and \(1.673 \times 10^{-27}\) kg).
02

Uncertainty Principle application

Use the Heisenberg Uncertainty Principle formula, \( \Delta x \Delta p \geq \hbar / 2 \), where \( \Delta x \) is the uncertainty in position (given as the diameter of nucleus here), \( \Delta p \) the uncertainty in momentum (which equals to \(m \Delta v \)), and \( \hbar = h / (2\pi) \) is the reduced Planck constant.
03

Minimum Speed Calculation for Electron and Proton

Rearrange the formula for \( \Delta v \) (uncertainty in speed) and apply the values for the electron and proton respectively. \( \Delta v \) = \( \hbar / (2m \Delta x) \)
04

Comparing to Light Speed

Compare the calculated speed of both proton and electron to the speed of light, and deduce from this comparison why an electron cannot be confined within a nucleus but a proton can.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 student placed a large unwrapped chocolate bar in a microwave oven without a rotating glass plate. After turning the oven on for less than a minute, she noticed there were evenly spaced dents (due to melting) about \(6 \mathrm{~cm}\) apart. Based on her observations, calculate the speed of light given that the microwave frequency is \(2.45 \mathrm{GHz}\). (Hint: The energy of a wave is proportional to the square of its amplitude.)

Which of the following species has the most unpaired electrons: \(\mathrm{S}^{+}, \mathrm{S},\) or \(\mathrm{S}^{-} ?\) Explain how you arrive at your answer.

What is electron configuration? Describe the roles that the Pauli exclusion principle and Hund's rule play in writing the electron configuration of elements.

Use the Aufbau principle to obtain the ground-state electron configuration of selenium.

What is the de Broglie wavelength, in centimeters, of a 12.4-g hummingbird flying at \(1.20 \times 10^{2} \mathrm{mph} ?\) \((1 \mathrm{mile}=1.61 \mathrm{~km})\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

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.

Sign-up for free