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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.

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