Chapter 7, Problem 138AP

The ionization energy of a certain element is 412 kJ/mol. When the atoms of this element are in the first excited state, however, the ionization energy is only 126 kJ/mol. Based on this information, calculate the wavelength of light emitted in a transition from the first excited state to the ground state.

Interpretation Introduction

Interpretation: The wavelength of light discharged in a transition from first state to a ground state is to be calculated.

Concept introduction:

Ionization energy is defined as the energy which is required to remove valence electron from a neutral atom or molecule in gaseous phase.

The ionization energy is expressed in $\text{kJ}/\text{mol}$.

The wavelength of light is represented as follows:

$\lambda =\frac{hc}{\Delta E}$

Here, $\Delta E$

is the energy difference, $h$

is Planck’s constant, and $c$

is the speed of light.

Answer to Problem 138AP

Solution: $419nm$

Explanation of Solution

Given information: ${\text{E}}_1=412kJ/mol$

${\text{E}}_2=126kJ/mol$

The energy difference between the ground state and the dissociation limit $\left(E_1\right)$ is equal to $412kJ/mol$, the energy difference between the first excited state and the dissociation limit $\left(E_2\right)$

is equal to $126kJ/mol$. So, the energy difference between the ground state and the excited state $\left(\Delta E\right)$ is given as follows:

$\Delta E=E_1-E_2$ …… (1)

Substitute $412kJ/mol$

for $E_1$ and $126kJ/mol$

for $E_2$

in equation (1) as follows,

$\begin{array}{c}\Delta E = 412\text{ kJ}/\text{mol} -126\text{ kJ}/\text{mol} \\ = 286\text{ kJ}/\text{mol}\end{array}$

In one kilojoule, there are ${10}^3$

joule present.

So,

$\begin{array}{c}286\text{ kJ}/\text{mol}=\left(286\frac{\cancel{\text{kJ}}}{\text{mol}}\right)\left({10}^3\frac{\text{J}}{\cancel{\text{kJ}}}\right)\\ =286\times {10}^3\text{ J}/\text{mol}\end{array}$

In one mole, there are $6.022\times {10}^{23}\text{photons}$ present.

So, in $286\times {10}^3\text{ J}/\text{mol}$, the number of photons are as follows:

$\left(\frac{\text{286}\text{×}{\text{10}}^{\text{3}}\text{J}}{\cancel{1\text{mol}}}\right)\times \left(\frac{\cancel{1\text{mol}}}{6.022\times {10}^{23}\text{photons}}\right)=4.75\times {10}^{-19}\text{J/photon}$

The wavelength of light that is emitted in a transition is calculated as follows:

$\lambda =\frac{hc}{\Delta E}$ …… (2)

Substitute $6.626\times {10}^{-34}\text{J}\text{.s}$ for $h$, $3.0\times {10}^8\text{m/s}$ for $c$, and $4.75\times {10}^{-19}\text{J/photon}$

for $\Delta E$

in equation (2) as follows:

$\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\cancel{\text{J}}\text{.}\cancel{\text{s}}\right)\left(3.0\times {10}^8\text{m/}\cancel{\text{s}}\right)}{\left(4.75\times {10}^{-19}\cancel{\text{J}}\right)}\\ =\text{419}\text{×}{\text{10}}^{\text{-9}}\text{m}\end{array}$

Since $1\text{ m}=\frac{1}{{\text{10}}^{-9}}\text{nm}$,

Therefore,

$\begin{array}{c}\text{419}\text{×}{\text{10}}^{\text{-9}}\text{m}=\left(\text{419}\text{×}{\text{10}}^{\text{-9}}\cancel{\text{m}}\right)\left(\frac{1}{{\text{10}}^{-9}}\text{nm/}\cancel{\text{m}}\right)\\ =\text{419}\text{ nm}\end{array}$

Conclusion

The wavelength of light discharged in a transition is $419\text{ nm}$.