Question

Quantum mechanics predicts that the energy of the ground state of the $\mathbf{H}$atom is $\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{eV}$. Insight into the magnitude of this quantity is gained by considering several methods by which it can be measured.

(a) Calculate the longest wavelength of light that will ionize$\mathbf{H}$atoms in their ground state.

(b) Assume the atom is ionized by collision with an electron that transfers all its kinetic energy to the atom in the ionization process. Calculate the speed of the electron before the collision. Express your answer in meters per second $\left({\mathrm{ms}}^{-1}\right)$and miles per hour (miles${\mathbf{h}}^{\mathbf{-}\mathbf{1}}$).

(c) Calculate the temperature required to ionize a $\mathbf{H}$atom in its ground state by thermal excitation. (Hint: Recall the criterion for thermal excitation of an oscillator in Planck's theory of blackbody radiation is that$\mathbf{h\nu }\mathbf{\approx }{\mathbf{k}}_{\mathbf{B}}\mathbf{T}$.)

Short answer

(a) The longest wavelength of light that can ionize the hydrogen atom in its ground state is $9.1161\times {10}^1\mathrm{nm}$.

(b) The speed of the electron before the collision is $4.7841\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

(c) The temperature required to ionize hydrogen atoms in their ground state by thermal excitation is $157824.3275\mathrm{K}$.

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:

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

where

- IE is ionization energy.

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

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

-${\mathit{\nu }}_{\mathbf{electron}}$is electron frequency.

(a): Calculating the longest wavelength of light that can ionize the hydrogen atom in its ground state

It is given that the energy of the ground-state hydrogen atom is $-13.6\mathrm{eV}$.

The threshold energy required to ionize the hydrogen atom in its ground state is $-13.6\mathrm{eV}$that is equal to $2.179\times {10}^{-18}\mathrm{J}$.

The formula for the wavelength of photon radiation is as follows:

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{E}}$

where

- $\mathit{\lambda }$is the wavelength of radiation.

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

- $\mathit{E}$is the energy of radiation.

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

The value of $E$is $2.179\times {10}^{-18}\mathrm{J}$.

The value of $c$is $2.9979\times {10}^{8}m·s^{-1}$.

The value of $h$is $6.626\times {10}^{-34}\mathrm{J}·{\mathrm{s}}^{-1}$.

Substitute these values in the equation above to calculate the wavelength of photon radiation.

$$\begin{gathered} \lambda =\frac{hc}{E} \\ \lambda =\frac{\left(6.626\times {10}^{-34}\mathrm{J}·{\mathrm{s}}^{-1}\right)\left(2.9979\times {10}^8\mathrm{m}·{\mathrm{s}}^{-1}\right)}{2.179\times {10}^{-18}\mathrm{J}} \\ \lambda =9.1161\times {10}^{-8}\mathrm{m}\left(\frac{{10}^9\mathrm{nm}}{1\mathrm{m}}\right) \\ \mathrm{\lambda }=9.1161\times {10}^1\mathrm{nm} \end{gathered}$$

The longest wavelength of light that can ionize the hydrogen atom in its ground state is $9.1161\times {10}^1\mathrm{nm}$.

(b): Calculating the speed of the electron before the collision

It is given that the energy of the ground state hydrogen atom is $-13.6\mathrm{eV}$.

Since an electron on collision transfers its kinetic energy to the hydrogen atom for ionization, the kinetic energy of the electron is equal to the threshold energy required for the ionization of the hydrogen atom, $-13.6\mathrm{eV}$, that is equal to $2.179\times {10}^{-18}\mathrm{J}$.

The formula for kinetic energy is as follows:

$\mathbf{KE}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}_{\mathbf{e}}{{\mathbf{v}}^{\mathbf{2}}}_{\mathbf{electron}}$

Rearrange the equation above to calculate ${\mathrm{v}}_{\mathrm{electron}}$, that is, the speed of the electron before the collision.

${\mathrm{v}}_{\mathrm{electron}}=\sqrt{\frac{\mathrm{KE}}{{\displaystyle \frac{1}{2}}{\mathrm{m}}_{\mathrm{e}}}}$

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

The value of $\mathrm{KE}$is $2.179\times {10}^{-18}\mathrm{J}$.

Substitute these values in the equation above to calculate ${\mathrm{v}}_{\mathrm{electron}}$.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{electron}}=\sqrt{\frac{2.179\times {10}^{-18}\mathrm{J}\left({\displaystyle \frac{1\mathrm{kg}·{\displaystyle \raisebox{1ex}{${\mathrm{m}}^2$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right.}}{1\mathrm{J}}}\right)}{{\displaystyle \frac{1}{2}}\left(9.1093\times {10}^{-31}\mathrm{kg}\right)}} \\ {\mathrm{v}}_{\mathrm{electron}}=4.7841\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right. \end{gathered}$$

The speed of electrons before the collision is $4.7841\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

(c): Calculating the temperature required to ionize hydrogen atom in its ground state by thermal excitation

The energy of the ground state hydrogen atom is $-13.6\mathrm{eV}$.

The criteria for thermal excitation of oscillation as per plank theory are as follows:

$\mathbf{h\nu }\mathbf{=}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}$

where

- $\mathbf{\nu }$is the frequency of radiation.

- ${\mathbf{k}}_{\mathbf{B}}$is the Boltzmann constant.

- $\mathbf{T}$is the temperature.

Rearrange the above equation to calculate $\left(T\right)$temperature required to ionize hydrogen atoms in their ground state by thermal excitation.

$\mathrm{T}=\frac{\mathrm{h\nu }}{{\mathrm{k}}_{\mathrm{B}}}$

The value of $\mathrm{h\nu }$is $-13.6\mathrm{eV}$that is equal to $2.179\times {10}^{-18}\mathrm{J}$.

The value of $k_B$is $1.380649\times {10}^{-23}\mathrm{J}·{\mathrm{K}}^{-1}$.

Substitute these values in the above equation to calculate $T$.

$$\begin{gathered} T=\frac{2.179\times {10}^{-18}\mathrm{J}}{1.380649\times {10}^{-23}\mathrm{J}·{\mathrm{K}}^{-1}} \\ T=157824.3275\mathrm{K} \end{gathered}$$

The temperature required to ionize hydrogen atoms in their ground state by thermal excitation is $157824.3275\mathrm{K}$.