Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 28: Problem 27

What is the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus \((1.0 \mathrm{fm}) ?\)

Short Answer

Expert verified
Answer: The minimum kinetic energy of an electron confined to a region the size of an atomic nucleus (1.0 fm) is approximately 1.763 × 10^(-14) J or 17.63 MeV.

Step by step solution

01

Calculate uncertainty in momentum (∆p)

Using Heisenberg Uncertainty Principle, ∆x * ∆p ≥ h/(4π) Given: ∆x = 1.0 fm = 1.0 × 10^(-15) m We want to find the minimum possible uncertainty in momentum (∆p), so we must find the equality. ∆p = (h/(4π))/ ∆x Now, substitute the known values and the given ∆x value: ∆p = ((6.626 × 10^(-34) Js)/(4π)) / (1.0 × 10^(-15) m)
02

Calculate the minimum kinetic energy (K.E.)

The formula for kinetic energy is: K.E. = p^2/(2m) The minimum kinetic energy will occur when the momentum (p) equals the minimum possible uncertainty in momentum (∆p). So, we can rewrite the formula as: K.E_min = (∆p)^2 / (2m) Substitute the value of ∆p from Step 1 and the mass of the electron (m = 9.11 × 10^(-31) kg): K.E_min = ((6.626 × 10^(-34) Js / (4π))/(1.0 × 10^(-15) m))^2 / (2 * (9.11 × 10^(-31) kg))
03

Calculate the numerical value of the minimum kinetic energy

Now, we can plug in the numbers and calculate the minimum kinetic energy. K.E_min = ((6.626 × 10^(-34) Js / (4π))/(1.0 × 10^(-15) m))^2 / (2 * (9.11 × 10^(-31) kg)) K.E_min ≈ 1.763 × 10^(-14) J Thus, the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus (1.0 fm) is approximately 1.763 × 10^(-14) J or 17.63 MeV.

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

Many lasers, including the helium-neon, can produce beams at more than one wavelength. Photons can stimulate emission and cause transitions between the \(20.66-\mathrm{eV}\) metastable state and several different states of lower energy. One such state is 18.38 eV above the ground state. What is the wavelength for this transition? If only these photons leave the laser to form the beam, what color is the beam?
An x-ray diffraction experiment using 16 -keV x-rays is repeated using electrons instead of \(x\) -rays. What should the kinetic energy of the electrons be in order to produce the same diffraction pattern as the x-rays (using the same crystal)?
(a) Show that the ground-state energy of the hydrogen atom can be written \(E_{1}=-k e^{2} /\left(2 a_{0}\right),\) where \(a_{0}\) is the Bohr radius. (b) Explain why, according to classical physics, an electron with energy \(E_{1}\) could never be found at a distance greater than \(2 a_{0}\) from the nucleus.
What is the largest number of electrons with the same pair of values for \(n\) and \(\ell\) that an atom can have?
List the number of electron states in each of the subshells in the \(n=7\) shell. What is the total number of electron states in this shell?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Electric Charge Field and Potential

Read Explanation

Classical Mechanics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Particle Model of Matter

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