Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 27: Problem 35

Find the energy in electron-volts required to remove the remaining electron from a doubly ionized lithium (Li \(^{2+}\) ) atom.

Short Answer

Expert verified
Answer: To completely remove the remaining electron from a doubly ionized lithium atom (Li \(^{2+}\)), it requires +122.4 electron-volts of energy.

Step by step solution

01

Identify the atomic number and energy level number for Li \(^{2+}\)

The atomic number Z of lithium (Li) is 3 because it has 3 protons in its nucleus. Since Li \(^{2+}\) is doubly ionized, only one electron remains and it is in the innermost shell, corresponding to energy level n = 1.
02

Apply the ionization energy formula

Using the ionization energy formula \(E = -\dfrac{13.6 \times Z^2}{n^2}\), we can calculate the energy required to remove the remaining electron from Li \(^{2+}\). Plug in the values for Z and n: \(E = -\dfrac{13.6 \times 3^2}{1^2}\).
03

Perform the calculations and express the result in electron-volts

Calculate the energy by plugging the values into the equation: \(E = -\dfrac{13.6 \times 9}{1} = -122.4\) eV. Since the energy is negative, it indicates that the energy is required to remove the electron. Thus, to ionize the Li \(^{2+}\) completely, we need +122.4 electron-volts of energy.

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 \(200-\) W infrared laser emits photons with a wavelength of $2.0 \times 10^{-6} \mathrm{m},\( and a \)200-\mathrm{W}$ ultraviolet light emits photons with a wavelength of \(7.0 \times 10^{-8} \mathrm{m}.\) (a) Which has greater energy, a single infrared photon or a single ultraviolet photon? (b) What is the energy of a single infrared photon and the energy of a single ultraviolet photon? (c) How many photons of each kind are emitted per second?
One line in the helium spectrum is bright yellow and has the wavelength $587.6 \mathrm{nm} .$ What is the difference in energy (in electron-volts) between two helium levels that produce this line?
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?
A Compton scattering experiment is performed using an aluminum target. The incident photons have wavelength \(\lambda\). The scattered photons have wavelengths \(\lambda^{\prime}\) and energies \(E\) that depend on the scattering angle \(\theta.\) (a) At what angle \(\theta\) are scattered photons with the smallest energy detected? (b) At this same scattering angle \(\theta,\) what is the ratio \(\lambda^{\prime} / \lambda\) for \(\lambda=10.0 \mathrm{pm} ?\)
What is the energy of a photon of light of wavelength \(0.70 \mu \mathrm{m} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Nuclear Physics

Read Explanation

Scientific Method Physics

Read Explanation

Famous Physicists

Read Explanation

Radiation

Read Explanation

Kinematics Physics

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