Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 53

Calculate the de Broglie wavelength for each of the following. a. an electron with a velocity \(10 . \%\) of the speed of light b. a tennis ball \((55 \mathrm{g})\) served at \(35 \mathrm{m} / \mathrm{s}(\sim 80 \mathrm{mi} / \mathrm{h})\)

Short Answer

Expert verified
The de Broglie wavelength for the electron with a velocity of 10% the speed of light is approximately \(2.426 × 10^{-12} m\). The de Broglie wavelength for the tennis ball served at 35 m/s is approximately \(3.445 × 10^{-34} m\).

Step by step solution

01

Calculate the electron's momentum

The given velocity of the electron is 10% of the speed of light. To find the actual velocity value, multiply the speed of light (c) by 0.1 (10%). The mass of an electron (m_e) is approximately \( 9.109 \times 10^{-31}\ kg \). So, \(v = 0.1c\) The electron's momentum (p_e) can be calculated as: \( p_e = m_ev = m_e(0.1c) \) Now, plug in the values for \(m_e\) and \(c\): \( p_e = (9.109 \times 10^{-31}\ kg)(0.1 \times 3.0 \times 10^{8}\ m/s) \) Calculate the momentum: \( p_e ≈ 2.7327 × 10^{-23} kg*m/s \)
02

Calculate the tennis ball's momentum

The mass of the tennis ball is given as 55g, which we need to convert to kilograms. The velocity of the tennis ball is given as 35 m/s. So, \(m_t = 55g = 0.055kg\), and \(v_t = 35 m/s\) The tennis ball's momentum (p_t) can be calculated as: \( p_t = m_tv_t \) Now, plug in the values for \(m_t\) and \(v_t\): \( p_t = (0.055\ kg)(35\ m/s) \) Calculate the momentum: \( p_t ≈ 1.925 kg*m/s \)
03

Calculate the de Broglie wavelengths

Now that we have the momentums for both the electron and the tennis ball, we can use the de Broglie equation to calculate their respective wavelengths. The Planck constant (h) is approximately \( 6.626 \times 10^{-34} Js \). For the electron: \( λ_e = \dfrac{h}{p_e} \) \( λ_e = \dfrac{6.626 \times 10^{-34} Js}{2.7327 × 10^{-23} kg*m/s} \) Calculate the wavelength: \( λ_e ≈ 2.426 × 10^{-12} m \) For the tennis ball: \( λ_t = \dfrac{h}{p_t} \) \( λ_t = \dfrac{6.626 \times 10^{-34} Js}{1.925 kg*m/s} \) Calculate the wavelength: \( λ_t ≈ 3.445 × 10^{-34} m \)
04

Final Results

The de Broglie wavelength for the electron with a velocity of 10% the speed of light is approximately \(2.426 × 10^{-12} m\). The de Broglie wavelength for the tennis ball served at 35 m/s is approximately \(3.445 × 10^{-34} m\).

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

A particle has a velocity that is \(90 . \%\) of the speed of light. If the wavelength of the particle is \(1.5 \times 10^{-15} \mathrm{m},\) what is the mass of the particle?

Assume that a hydrogen atom's electron has been excited to the \(n=6\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

What is the physical significance of the value of \(\psi^{2}\) at a particular point in an atomic orbital?

How many valence electrons do each of the following elements have, and what are the specific valence electrons for each element? a. Ca b. O c. element 117 d. In e. Ar f. Bi

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and \(589.6 \mathrm{nm}\). What are the frequency and the energy of a photon of light at each of these wavelengths? What are the energies in \(\mathrm{kJ} / \mathrm{mol} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

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