Chapter 8, Problem 8.102QP

Interpretation Introduction

Interpretation:

The wavelength required to remove one electron from lithium atom should be determined.

Concept Introduction:

Atomic Number: Atomic number of the element is equal to the number of protons present in the nucleus of the element which is denoted by symbol Z. The superscript presents on the left side of the symbol of the element.

First ionization energy:

The ionization energy is the minimum energy required to remove the electron from an isolated atom which is in the gaseous state results to give gaseous ion with one positive charge.

${\text{atom}}_{\left(\text{g}\right)}\to {\text{ion with positive charge}}_{\left(\text{g}\right)}\text{ + electron}$

Cation: Removal of electron from the atom results to form positively charged ion called cation.

The net charge present in the element denotes the presence or absence of electrons in the element.

In periodic table the horizontal rows are called periods and the vertical column are called group.

Plank-Einstein Equation: The energy is conversed property since it can neither be created nor be destroyed but can be transformed. The energy of the photon is obtained by using the following relation

$\begin{array}{l}\text{E}\text{=}\text{hν}\\ \text{h}\to \text{Planck's constant = 6}\text{.63}\times {\text{10}}^{\text{-34}}\text{J}\text{.s}\\ \text{ν}\to \text{frequency}\text{of}\text{the}\text{photon}\end{array}$

Wavelength: The distance between the two continuous maximum displacements present in wave or the two continuous minimum displacements present in a wave exhibited by the photons is called wavelength. The wavelength of the photon is inversely proportional to its frequency. The relationship between them is given by the following formula,

$\text{(wavelength)λ}\text{=}\frac{\text{(velocity of light)c}}{\text{(frequency)ν}}$

Frequency: It denotes the number of waves passes in given amount of time.

The wavelength in $\text{nm}$ for first ionization of lithium atom.

Answer to Problem 8.102QP

The wavelength required to remove the electron from the lithium atom is 230$\text{nm}$.

Explanation of Solution

Determine the ionization energy for lithium.

$\begin{array}{l}\text{Energy required to ionize one Li atom :}\\ \text{First ionization energy of Li atom = 520 kJ/mol}\\ =\frac{\text{520}\times {\text{10}}^{\text{3}}\text{J}}{1\text{mol}}\times \frac{1\text{mol}}{6.022\times {10}^{23}\text{atoms}}=8.635\times {10}^{-19}\text{J/atom}\end{array}$

The ionization energy for lithium atom is determined by using the given first ionization energy of the atom and converting it from $\text{kJ/mol}$ to $\text{J/atom}$ by dividing the ionization energy value with the number of atoms present.

Determine the wavelength required to remove electron from lithium atom.

$\begin{array}{l}\text{E}\text{=}\text{hν}\\ \text{(frequency)ν}\text{=}\frac{\text{(velocity of light)c}}{\text{(wavelength)λ}}\\ \text{E}\text{=}\text{h}\frac{\text{(velocity of light)c}}{\text{(wavelength)λ}}\\ \text{λ=}\text{h}\frac{\text{c}}{\text{E}}=\frac{\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)\left(3\times {10}^8\text{m/s}\right)}{8.635\times {10}^{-19}\text{J}}\\ =2.30\times {10}^{-7}\text{m}\\ =230\times {10}^{-2}\times {10}^{-7}\text{m}\\ \text{=230 nm }\left(\text{since}\text{1nm}\text{=}{\text{10}}^{\text{-9}}\text{m}\right)\end{array}$

The wavelength required to remove the electron from the nucleus is obtained by dividing Planck’s constant and velocity of light with the energy of the wavelength which yields the value in meters that is finally converted to nanometers using the conversion that $\text{1nm}\text{=}{\text{10}}^{\text{-9}}\text{m}$.

Conclusion

The maximum wavelength required to remove an electron from the atom is determined by using the ionization energy.