Chapter 5, Problem 83A

Interpretation Introduction

Interpretation : The energy required to eject an electron from a hydrogen atom when the electron is in the ground state and the second energy level is to be calculated. The energy required to eject a ground state electron from the lithium species is to be calculated.

Concept Introduction : The energy needed to remove an electron from an orbit of a hydrogen atom or an ion with only one electron can be calculated using Bohr's atomic theory. The ionization energy of the atom or ion is represented by a number. The formula of ionization energy is given as:

  $E=Z^2\times \frac{k}{n^2}$

Where, $Z$ is the atomic number, $k$ is $2.18\times {10}^{-18}\text{J}$ , and n is the energy level.

Answer to Problem 83A

The energy of hydrogen atom at ground state is $2.18\times {10}^{-18}\text{J}$ .

The energy of hydrogen atom at second energy state is $5.45\times {10}^{-19}\text{J}$ .

The energy of ${\text{Li}}^{\text{2+}}$ atom at ground state is $1.96\times {10}^{-17}\text{J}$ .

Explanation of Solution

The formula of ionization energy is given as:

  $E=Z^2\times \frac{k}{n^2}$

Where, $Z$ is the atomic number, $k$ is $2.18\times {10}^{-18}\text{J}$ , and n is the energy level.

The atomic number of a hydrogen atom is 1 and n is 1 since it is the ground state.

To calculate the energy for hydrogen atom at ground state (n=1), substitute the values in the above formula:

  $\begin{array}{c}E=Z^2\times \frac{k}{n^2}\\ E=1^2\times \frac{2.18\times {10}^{-18}\text{J}}{1^2}\\ =2.18\times {10}^{-18}\text{J}\end{array}$

If the hydrogen atom is in the second energy level (n=2), then the energy will be given as:

  $\begin{array}{c}E=Z^2\times \frac{k}{n^2}\\ E=1^2\times \frac{2.18\times {10}^{-18}\text{J}}{2^2}\\ =5.45\times {10}^{-19}\text{J}\end{array}$

For ${\text{Li}}^{\text{2+}}$ atom, the atomic number is 3 and after losing two electrons in it there is only one electron. The energy level is the ground state (n=1).

The energy is given as:

  $\begin{array}{c}E=Z^2\times \frac{k}{n^2}\\ E=3^2\times \frac{2.18\times {10}^{-18}\text{J}}{1^2}\\ =1.96\times {10}^{-17}\text{J}\end{array}$