Question

Photoelectron spectroscopy studies of sodium atoms excited by X-rays with wavelength$\mathbf{9}\mathbf{.}\mathbf{890}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathbf{m}$show four peaks in which the electrons have speeds $\mathbf{7}\mathbf{.}\mathbf{992}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$, role="math" localid="1663410225622" $\mathbf{2}\mathbf{.}\mathbf{046}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$, $\mathbf{2}\mathbf{.}\mathbf{074}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$and $\mathbf{2}\mathbf{.}\mathbf{009}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$. (Recall that$\mathbf{1}\mathbf{}\mathbf{J}\mathbf{=}\mathbf{1}\mathbf{}{\mathbf{kgm}}^{\mathbf{2}}{\mathbf{s}}^{\mathbf{-}\mathbf{2}}$.)

(a) Calculate the ionization energy of the electrons in each peak.

(b) Assign each peak to an orbital of the sodium atom.

Short answer

(a) Ionization energy of an electron in each peak is $1.7175\times {10}^{-16}\mathrm{J},1.01840\times {10}^{-17}\mathrm{J},4.93013\times {10}^{-18}\mathrm{J}$and $1.70176\times {10}^{-17}\mathrm{J}$.

(b) Peak corresponds to energy $4.93013\times {10}^{-18}\mathrm{J}$represent the removal of electrons from $3\mathrm{s}$which are least tightly held.

role="math" localid="1663410662144" $1.7175\times {10}^{-16}\mathrm{J}$depict the ejection of an electron from $2{\mathrm{p}}^6$.

$1.70176\times {10}^{-17}\mathrm{J}$represent the removal of an electron from $2{\mathrm{p}}^5$and $1.01840\times {10}^{-17}\mathrm{J}$removal of an electron from $2{\mathrm{p}}^4$.

Step-by-step solution

Concept and formula used to calculate the ionization energy of an electron

The shell structure is confirmed by photoelectron spectroscopy or (PES).

Photoelectron spectroscopy indicates orbital energy states by evaluating the ionization energy required to remove electrons from the atom.

The formula for ionization energy is as follows:

$\mathbf{IE}\mathbf{=}{\mathbf{h\nu }}_{\mathbf{photon}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}_{\mathbf{e}}{{\mathbf{v}}^{\mathbf{2}}}_{\mathbf{electron}}$

where

- IE is ionization energy.

- $\mathbf{h}$is planks constant.

- ${\mathbf{\nu }}_{\mathbf{photon}\mathbf{}\mathbf{us}\mathbf{}\mathbf{photon}\mathbf{}\mathbf{frequency}}\mathbf{-}{\mathbf{m}}_{\mathbf{e}}$ is the mass of an electron.

- ${\mathbf{v}}_{\mathbf{electron}}$is electron frequency.

(a): Calculation of the ionization energy in peak when electron speed is 7.992×106 m·s-1

Sodium atom is excited by X-ray with wavelength $9.890\times {10}^{-10}\mathrm{m}$and the 4 peaks are obtained with a speed of electron as $7.992\times {10}^6\mathrm{m}·{\mathrm{s}}^{-1},2.046\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1},2.074\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$and $2.009\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$.

The formula for ionization energy is as follows:

$\mathrm{IE}={\mathrm{h\nu }}_{\mathrm{photon}}-\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{{\mathrm{v}}^2}_{\mathrm{electron}}$ …… (1)

The formula to calculate the frequency of X-ray is as follows:

${\mathbf{\nu }}_{\mathbf{X}\mathbf{-}\mathbf{ray}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{\lambda }}$ …… (2)

where

- $\mathit{c}$is the speed of light

- $\mathit{\lambda }$is the wavelength of the X-ray incident

Value of $c$is role="math" localid="1663412234554" $2.9979\times {10}^8\mathrm{m}·{\mathrm{s}}^{-1}$

Value of $\lambda$is $9.890\times {10}^{-10}\mathrm{m}$.

Substitute these values in equation (2) to calculate ${\mathrm{\nu }}_{\mathrm{X}-\mathrm{ray}}$.

$$\begin{gathered} {\mathrm{\nu }}_{\mathrm{X}-\mathrm{ray}}=\frac{\mathrm{c}}{\mathrm{\lambda }} \\ {\mathrm{\nu }}_{\mathrm{X}-\mathrm{ray}}=\frac{2.9979\times {10}^8\mathrm{m}·{\mathrm{s}}^{-1}}{9.980\times {10}^{-10}\mathrm{m}} \\ {\mathrm{\nu }}_{\mathrm{X}-\mathrm{ray}}=3.0312\times {10}^{17}{\mathrm{s}}^{-1} \end{gathered}$$

Value of $\mathrm{\nu }$X-ray is $3.0312\times {10}^{17}{\mathrm{s}}^{-1}$.

Value of ${\mathrm{m}}_{\mathrm{e}}$is $9.1093\times {10}^{-31}\mathrm{kg}$.

Value of $\mathrm{h}$is $6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}$.

Value of ${\mathrm{\nu }}_{\mathrm{electron}}$is role="math" localid="1663412838640" $7.992\times {10}^6\mathrm{m}·{\mathrm{s}}^{-1}$.

Substitute these values in equation (1) to calculate the ionization energy of peak with electron speed $7.992\times {10}^6\mathrm{m}·{\mathrm{s}}^{-1}$.

$$\begin{gathered} \mathrm{IE}={\mathrm{hv}}_{\mathrm{X}-\mathrm{ray}}-\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{{\mathrm{\nu }}_{\mathrm{electron}}}^2 \\ \mathrm{IE}=\left(6.626\times {10}^{-34}\mathrm{J}\left(\frac{1\mathrm{kg}·{\displaystyle \raisebox{1ex}{${\mathrm{m}}^2$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right.}}{1\mathrm{J}}\right)·\mathrm{s}\right)\left(3.0312\times {10}^{17}{\mathrm{s}}^{-1}\right)-\frac{1}{2}\left(9.1093\times {10}^{-31}\mathrm{kg}\right)\left(7.992\times {10}^{17}{\mathrm{s}}^{-1}\right) \\ \mathrm{IE}=1.7175\times {10}^{-16}\mathrm{J} \end{gathered}$$

Step 3: Calculation of the ionization energy in peak when electron speed is 2.046×107 m·s-1

Value of ${\mathrm{\nu }}_{\mathrm{X}-\mathrm{ray}}$is $3.0312\times {10}^{17}{\mathrm{s}}^{-1}$.

Value of ${\mathrm{m}}_{\mathrm{e}}$is $9.1093\times {10}^{-31}\mathrm{kg}$.

Value of $h$is $6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}$.

Value of ${\mathrm{\nu }}_{\mathrm{electron}}$is $2.046\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$.

Substitute these values in equation (1) to calculate ionization energy of peak with electron speed $2.046\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$

$$\begin{gathered} \mathrm{IE}={\mathrm{h\nu }}_{\mathrm{X}-\mathrm{ray}}-\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{{\mathrm{\nu }}_{\mathrm{electron}}}^2 \\ \mathrm{IE}=\left(\left(6.626\times {10}^{-34}\mathrm{J}\left(\frac{1\mathrm{kg}·{\displaystyle \raisebox{1ex}{${\mathrm{m}}^2$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right.}}{1\mathrm{J}}\right)·\mathrm{s}\right)\left(3.0312\times {10}^{17}{\mathrm{s}}^{-1}\right)\right)-\frac{1}{2}\left(9.1093\times {10}^{-31}\mathrm{kg}\right){\left(2.046\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}\right)}^2 \\ \mathrm{IE}=1.01840\times {10}^{-17}\mathrm{J} \end{gathered}$$

Step 4: Calculation of the ionization energy in peak when electron speed is 2.074×107 m·s-1

Value of${\mathrm{\nu }}_{\mathrm{X}-\mathrm{ray}}$is $3.0312\times {10}^{17}{\mathrm{s}}^{-1}$.

Value of ${\mathrm{m}}_{\mathrm{e}}$is $9.1093\times {10}^{-31}\mathrm{kg}$.

Value of $h$is $6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}$.

Value of ${\mathrm{\nu }}_{\mathrm{electron}}$is $2.074\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$.

Substitute these values in equation (1) to calculate ionization energy of peak with electron speed$2.074\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$

$$\begin{gathered} \mathrm{IE}={\mathrm{h\nu }}_{\mathrm{X}-\mathrm{ray}}-\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{{\mathrm{\nu }}_{\mathrm{electron}}}^2 \\ \mathrm{IE}=\left(\left(6.626\times {10}^{-34}\mathrm{J}\left(\frac{1\mathrm{kg}·{\displaystyle \raisebox{1ex}{${\mathrm{m}}^2$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right.}}{1\mathrm{J}}\right)·\mathrm{s}\right)\left(3.0312\times {10}^{17}{\mathrm{s}}^{-1}\right)\right)-\frac{1}{2}\left(9.1093\times {10}^{-31}\mathrm{kg}\right){\left(2.074\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}\right)}^2 \\ \mathrm{IE}=4.93013\times {10}^{-18}\mathrm{J} \end{gathered}$$

Step 5: Calculation of the ionization energy in peak when electron speed is 2.009×107 m·s-1

Value of ${\mathrm{\nu }}_{\mathrm{X}-\mathrm{ray}}$is $3.0312\times {10}^{17}{\mathrm{s}}^{-1}$

Value of ${\mathrm{m}}_{\mathrm{e}}$is $9.1093\times {10}^{-31}\mathrm{kg}$.

Value of $h$is $6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}$.

Value of ${\mathrm{\nu }}_{\mathrm{electron}}$is $2.009\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$

Substitute these values in equation (1) to calculate the ionization energy of peak with electron speed $2.009\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$.$\begin{array}{l}\mathrm{IE}=h{\nu }_{\mathrm{X}-\mathrm{ray}}-\frac{1}{2}{m}_e{v_{\mathrm{electron}}}^2\\ =\left(\left(6.626\times {10}^{-34}\mathrm{J}\left(\frac{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}{1\mathrm{J}}\right)·\mathrm{s}\right)\left(3.0312\times {10}^{17}{\mathrm{s}}^{-1}\right)\right)-\frac{1}{2}\left(9.1093\times {10}^{-31}\mathrm{kg}\right){\left(2.009\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}\right)}^2\\ =1.70176\times {10}^{-17}\mathrm{J}\end{array}$

(b): Explanation for assigning each peak to the orbital of the sodium atom

Sodium atom is excited by $\mathrm{X}$ray with wavelength $9.890\times {10}^{-10}\mathrm{m}$and the 4 peaks are obtained with a speed of electron as $7.992\times {10}^6\mathrm{m}·{\mathrm{s}}^{-1},2.046\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1},2.074\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$and $2.009\times {10}^7\mathrm{m}·{\mathrm{s}}^{-1}$.

The ionization energy of electrons in each peak is $1.7175\times {10}^{-16}\mathrm{J},1.01840\times {10}^{-17}\mathrm{J},4.93013\times {10}^{-18}\mathrm{J}$and $1.70176\times {10}^{-17}\mathrm{J}$.

The electronic configuration of the sodium is $1s^{2}2s^{2}2p^{6}3s^1$.

The peak corresponds to energy$4.93013\times {10}^{-18}\mathrm{J}$ represent the removal of electrons from $3s$which are least tightly held.

The energy required to remove from the noble configuration is higher than any other configuration so $1.7175\times {10}^{-16}\mathrm{J}$represent the removal of an electron from $2p^6$.

$1.70176\times {10}^{-17}\mathrm{J}$represent the removal of an electron from $2p^5$and $1.01840\times {10}^{-17}\mathrm{J}$removal of an electron from $2p^4$as this leads to the formation of a half-filled configuration.