Chapter 11, Problem 1ASA

(a)

Interpretation Introduction

Interpretation:

The energies in kJ/mol for four lowest energy levels of the ${\text{Be}}^{\text{+3}}$ ions bromine must be calculated.

Concept introduction: Bohr’s theory of atoms was applicable to the one electron system. It purposed the energy levels in an atom. It purposed the mathematical relation for the calculation of the radius of orbits, energy and velocity of electrons.

  $\text{Energy = }\frac{\text{-13}{\text{.6Z}}^{\text{2}}}{{\text{n}}^{\text{2}}}\text{eV = }\frac{\text{-2}{\text{.179 × 10}}^{\text{-21}}{\text{ ×Z}}^{\text{2}}}{{\text{n}}^{\text{2}}}\text{ kJ=}\frac{\text{-21001 }}{{\text{n}}^{\text{2}}}\text{kJ/mol}$

Answer to Problem 1ASA

  $\begin{array}{l}{\text{E}}_{\text{1 }}\text{=}\frac{\text{-21001 }}{{\text{(1)}}^{\text{2}}}\text{kJ/mol = -21001 kJ/mol}\\ {\text{E}}_{\text{2 }}\text{=}\frac{\text{-21001 }}{{\text{(2)}}^{\text{2}}}\text{kJ/mol = -5250 kJ/mol}\\ {\text{E}}_{\text{3 }}\text{=}\frac{\text{-21001 }}{{\text{(3)}}^{\text{2}}}\text{kJ/mol = -2333 kJ/mol}\\ {\text{E}}_{\text{4 }}\text{=}\frac{\text{-21001 }}{{\text{(4)}}^{\text{2}}}\text{kJ/mol = -1,312}\text{.5 kJ/mol}\end{array}$

Explanation of Solution

The atomic number of Beryllium = 4

Thus, the electronic configuration must be= ${\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}$ .

For the first four energy levels, the value of $\text{n}$ must be $\text{1, 2, 3 and 4}$ . Substitute the values to calculate the energy:

  $\text{Energy =}\frac{\text{-21001 }}{{\text{n}}^{\text{2}}}\text{kJ/mol}$

  $\begin{array}{l}{\text{E}}_{\text{1 }}\text{=}\frac{\text{-21001 }}{{\text{(1)}}^{\text{2}}}\text{kJ/mol = -21001 kJ/mol}\\ {\text{E}}_{\text{2 }}\text{=}\frac{\text{-21001 }}{{\text{(2)}}^{\text{2}}}\text{kJ/mol = -5250 kJ/mol}\\ {\text{E}}_{\text{3 }}\text{=}\frac{\text{-21001 }}{{\text{(3)}}^{\text{2}}}\text{kJ/mol = -2333 kJ/mol}\\ {\text{E}}_{\text{4 }}\text{=}\frac{\text{-21001 }}{{\text{(4)}}^{\text{2}}}\text{kJ/mol = -1,312}\text{.5 kJ/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation: The energies for the transition from n=2 to n=1 level for ${\text{Be}}^{\text{+3}}$ ions must be calculated along with the wavelength.

Concept introduction: Bohr’s theory of atoms was applicable to the one electron system. It purposed the energy levels in an atom. It purposed the mathematical relation for the calculation of the radius of orbits, energy and velocity of electrons.

  $\text{Energy = }\frac{\text{-13}{\text{.6Z}}^{\text{2}}}{{\text{n}}^{\text{2}}}\text{eV = }\frac{\text{-2}{\text{.179 × 10}}^{\text{-21}}{\text{ ×Z}}^{\text{2}}}{{\text{n}}^{\text{2}}}\text{ kJ=}\frac{\text{-21001 }}{{\text{n}}^{\text{2}}}\text{kJ/mol}$

Answer to Problem 1ASA

  $\Delta \text{E = 15,751 kJ/mol}$

  $\text{ λ=7}\text{.60nm}$

Explanation of Solution

The atomic number of Beryllium = 4

Thus, the electronic configuration must be= ${\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}$

For the first four energy levels, the value of $\text{n}$ must be $\text{1, 2, 3 and 4}$ . Substitute the values to calculate the energy:

  $\text{Energy =}\frac{\text{-21001 }}{{\text{n}}^{\text{2}}}\text{kJ/mol}$

The energies for the transition from n = 2 to n = 1 level for ${\text{Be}}^{\text{+3}}$ ions must be:

  $\Delta {\text{E = E}}_{\text{2 }}-{\text{E}}_{\text{1 }}\text{= ( -5250 kJ/mol})-(\text{ -21001 kJ/mol) = 15,751 kJ/mol}$

The wavelength will be:

  $\begin{array}{l}\text{ λ=}\frac{\text{hC}}{\text{E}}\\ \text{λ=}\frac{\text{6}{\text{.626×10}}^{\text{-34}}\text{J×3}{\text{.0×10}}^{\text{8}}\text{m/s×6}{\text{.022×10}}^{\text{23}}\text{ mol/atom}}{\text{15,751×10}\text{J}\text{/mol}}\text{= 7}{\text{.60×10}}^{\text{-9}}\text{m = 7}\text{.60 nm}\end{array}$

(c)

Interpretation Introduction

Interpretation: The quantum number of the initial and final states for the transition of given three wavelengths $\text{6}\text{.078 nm, 6}\text{.411 nm, 117}\text{.23 nm}$ must be calculated.

Concept introduction: Bohr’s theory of atoms was applicable to the one electron system. It purposed the energy levels in an atom. It purposed the mathematical relation for the calculation of the radius of orbits, energy and velocity of electrons.

  $\text{Energy = }\frac{\text{-13}{\text{.6Z}}^{\text{2}}}{{\text{n}}^{\text{2}}}\text{eV = }\frac{\text{-2}{\text{.179 × 10}}^{\text{-21}}{\text{ ×Z}}^{\text{2}}}{{\text{n}}^{\text{2}}}\text{ kJ=}\frac{\text{-21001 }}{{\text{n}}^{\text{2}}}\text{kJ/mol}$

Answer to Problem 1ASA

  a) 4 → 1 b) 3 → 1 c) 4 → 3

Explanation of Solution

Calculate energy fir the given three wavelengths $\text{6}\text{.078 nm, 6}\text{.411 nm, 117}\text{.23 nm}$ :

  $\begin{array}{l}{E}_1=\frac{hC}{{\lambda }_1}\\ E_1=\frac{6.626\times 10^{-34}J\times 3.0\times 10^{8}m/s\times 6.022\times 10^{23}}{6.078\times 10^{-9}m}=19.69\times 10^{6}J/mol=19,690kJ/mol\end{array}$

  $\begin{array}{l}{E}_2=\frac{hC}{{\lambda }_2}\\ E_2=\frac{6.626\times 10^{-34}J\times 3.0\times 10^{8}m/s\times 6.022\times 10^{23}}{6.411\times 10^{-9}m}=18.67\times 10^{6}J/mol=18672kJ/mol\end{array}$

  $\begin{array}{l}{E}_3=\frac{hC}{{\lambda }_3}\\ E_3=\frac{6.626\times 10^{-34}J\times 3.0\times 10^{8}m/s\times 6.022\times 10^{23}}{117.23\times 10^{-9}m}=1.02\times 10^{6}J/mol=1021kJ/mol\end{array}$

From the part (a):

  $\begin{array}{l}{\text{E}}_{\text{1 }}\text{= -21001 kJ/mol}\\ {\text{E}}_{\text{2 }}\text{= -5250 kJ/mol}\\ {\text{E}}_{\text{3 }}\text{= -2333 kJ/mol}\\ {\text{E}}_{\text{4 }}\text{= -1,312}\text{.5 kJ/mol}\end{array}$

Thus, the difference in ${\text{E}}_1$ and ${\text{E}}_4$ gives the transition energy exactly equal to the energy calculated from the ${\lambda }_1$ :

  $\Delta {\text{E}}_{\text{1}}{\text{=19690 = E}}_{\text{4}}{\text{ - E}}_{\text{1}}\text{ = -1312}\text{.5-(-21001)kJ/mol}$

Hence the ${\lambda }_1$ must be associated with transition from 4 → 1.

The difference in ${\text{E}}_1$ and ${\text{E}}_3$ gives the transition energy exactly equal to the energy calculated from the ${\lambda }_2$ . Hence the ${\lambda }_2$ must be associated with transition from 3 → 1.

The difference in ${\text{E}}_3$ and ${\text{E}}_4$ gives the transition energy exactly equal to the energy calculated from the ${\lambda }_3$ . Hence the ${\lambda }_3$ must be associated with transition from 4 → 3.