Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 28: Problem 9

What is the ratio of the wavelength of a 0.100-keV photon to the wavelength of a 0.100-keV electron?

Short Answer

Expert verified
The ratio of the wavelength of a 0.100-keV photon to the wavelength of a 0.100-keV electron is approximately 32:1.

Step by step solution

01

Planck's equation for photons

We will use Planck's equation to find the wavelength of the photon. Planck's equation relates the energy \(E_p\) of a photon to its wavelength \(\lambda_p\) and is given by: $$ E_p = h \frac{c}{\lambda_p} $$ where \(h\) is the Planck's constant (\(6.626 \times 10^{-34} Js\)) and \(c\) is the speed of light (\(3 \times 10^8 m/s\)).
02

de Broglie wavelength equation for electrons

For the electron, we'll use the de Broglie wavelength equation, which relates the energy \(E_e\) of an electron to its wavelength \(\lambda_e\). The equation is given by: $$ \lambda_e = \frac{h}{\sqrt{2 m_e E_e}} $$ where \(m_e\) is the mass of the electron (\(9.109 \times 10^{-31} kg\)).
03

Convert energy to Joules

Both energy values given are in keV, we need to convert them to Joules for further calculations. The conversion factor is \(1 keV = 1.602 \times 10^{-16} J\). Thus, $$ E_p = E_e = 0.1\text{ keV} \times 1.602 \times 10^{-16}J/keV = 1.602 \times 10^{-17}J $$
04

Calculate the photon's wavelength

Now, we can calculate the wavelength of the photon using Planck's equation: $$ \lambda_p = \frac{hc}{E_p} = \frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{1.602 \times 10^{-17}J} = 1.239 \times 10^{-9} m $$
05

Calculate the electron's wavelength

Next, we calculate the wavelength of the electron using de Broglie wavelength equation: $$ \lambda_e = \frac{h}{\sqrt{2 m_e E_e}} = \frac{6.626 \times 10^{-34} Js}{\sqrt{2 \times 9.109 \times 10^{-31} kg \times 1.602 \times 10^{-17} J}} = 3.870 \times 10^{-11} m $$
06

Find the ratio of the two wavelengths

Finally, we will find the ratio of the wavelength of the photon to the wavelength of the electron: $$ \frac{\lambda_p}{\lambda_e} = \frac{1.239 \times 10^{-9} m}{3.870 \times 10^{-11} m} \approx 32.03 $$ The ratio of the wavelength of a 0.100-keV photon to the wavelength of a 0.100-keV electron is approximately 32:1.

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

What is the ground-state electron configuration of bromine (Br, atomic number 35)?
What is the magnitude of the momentum of an electron with a de Broglie wavelength of \(0.40 \mathrm{nm} ?\)
At a baseball game, a radar gun measures the speed of a 144-g baseball to be \(137.32 \pm 0.10 \mathrm{km} / \mathrm{h} .\) (a) What is the minimum uncertainty of the position of the baseball? (b) If the speed of a proton is measured to the same precision, what is the minimum uncertainty in its position?
(a) What are the electron configurations of the ground states of lithium \((Z=3),\) sodium \((Z=11),\) and potassium \((Z=19) ?\) (b) Why are these elements placed in the same column of the periodic table?
A radar pulse has an average wavelength of \(1.0 \mathrm{cm}\) and lasts for \(0.10 \mu \mathrm{s}\). (a) What is the average energy of the photons? (b) Approximately what is the least possible uncertainty in the energy of the photons?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Work Energy and Power

Read Explanation

Measurements

Read Explanation

Rotational Dynamics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Oscillations

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