Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 27: Problem 49

By directly substituting the values of the fundamental constants, show that the Bohr radius \(a_{0}=\hbar^{2} /\left(m_{\mathrm{e}} k e^{2}\right)\) has the numerical value \(5.29 \times 10^{-11} \mathrm{m}.\)

Short Answer

Expert verified
Answer: The approximate numerical value of the Bohr radius is \(5.29 \times 10^{-11} m\).

Step by step solution

01

Identify the Constants

The variables in the given formula for the Bohr radius represent the following fundamental constants: - \(\hbar\) is the reduced Planck constant, with a value of \(1.0545718 \times 10^{-34} J\cdot s\) - \(m_e\) is the electron mass, with a value of \(9.10938356 \times 10^{-31} kg\) - \(k\) is the Coulomb's constant or electrostatic constant, with a value of \(8.98755179 \times 10^9 N\cdot m^2C^{-2}\) - \(e\) is the elementary charge (charge of an electron), with a value of \(1.60217662 \times 10^{-19} C\)
02

Substitute the Constants into the Formula

Now we can substitute the values of these constants into the formula for the Bohr radius: \(a_0 = \frac{\hbar^2}{(m_e\cdot k\cdot e^2)}\) \(a_0 = \frac{(1.0545718 \times 10^{-34})^2}{(9.10938356 \times 10^{-31}) \cdot (8.98755179 \times 10^9) \cdot (1.60217662 \times 10^{-19})^2}\)
03

Calculate the Numerical Value

Now that we have substituted the constants into the formula, we can calculate the numerical value of the Bohr radius: \(a_0 = \frac{(1.1111113 \times 10^{-68})}{(1.459762 \times 10^{-9})} \approx 5.291772 \times 10^{-11} m\) Thus, the Bohr radius has a numerical value of approximately \(5.29 \times 10^{-11} m\).

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

Show that the cutoff frequency for an x-ray tube is proportional to the potential difference through which the electrons are accelerated.
In a photoelectric experiment using sodium, when incident light of wavelength \(570 \mathrm{nm}\) and intensity \(1.0 \mathrm{W} / \mathrm{m}^{2}\) is used, the measured stopping potential is \(0.28 \mathrm{V} .\) (a) What would the stopping potential be for incident light of wavelength \(400.0 \mathrm{nm}\) and intensity \(1.0 \mathrm{W} / \mathrm{m}^{2} ?\) (b) What would the stopping potential be for incident light of wavelength 570 nm and intensity $2.0 \mathrm{W} / \mathrm{m}^{2} ?(\mathrm{c})$ What is the work function of sodium?
How much energy must be supplied to a hydrogen atom to cause a transition from the ground state to the \(n=4\) state?
What is the energy of a photon of light of wavelength \(0.70 \mu \mathrm{m} ?\)
What is the shortest wavelength \(\mathrm{x}\) -ray produced by a \(0.20-\mathrm{MV}\) x-ray machine?
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Nuclear Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Famous Physicists

Read Explanation

Electromagnetism

Read Explanation

Electricity and Magnetism

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