Chapter 8, Problem 8.1IA

(a)

Interpretation Introduction

Interpretation:

The transition of an electron form ground state ${\text{He}}^{\text{+}}$ to a state with quantum numbers, $n=4$, $l=1$, $m_l=+1$ has to be described using term symbols.

Concept introduction:

A term symbol describes the orbital, spin and total angular momenta of an electronic state.  The symbol $L$ represents the vector sum of orbital angular momentum.  The symbol $S$ represents the vector sum of spin angular momentum.  The symbol $J$ represents the total angular momentum.  Therefore, the term symbol is represented as given below.

  ${}^{2S+1}{L}_J$

Answer to Problem 8.1IA

The transition of an electron from ground state ${\text{He}}^{\text{+}}$ to a state with quantum numbers, $n=4$, $l=1$, $m_l=+1$ is represented as ${}^{\text{2}}{\text{S}}_{\text{1}/\text{2}}{\to }^{\text{2}}{\text{P}}_{\text{1}/\text{2}}$.

Explanation of Solution

The term symbol for the ${\text{He}}^{\text{+}}$ is written as shown below.

  ${}^{2S+1}{L}_J\mathrm{..............}\left(1\right)$

Where,

$L$ is the quantum number indicating sum of orbital angular momentum

$S$ is the spin angular momentum

$J$ is the total angular momentum.

The value of $L$ is equal to the value of $l$.

The value of $l$ in the excited state is $1$.

Therefore the value of $L$ is $1$ which corresponds to the term $\text{P}$.

The value of $S$ for ${\text{He}}^{\text{+}}$ ion in the excited state is $\frac{1}{2}$

Therefore, the value of $2S+1$ is calculated below.

    $\begin{array}{c}2S+1=2\times \frac{1}{2}+1\\ =2\end{array}$

As the orbital is less than half filled in excited state ${\text{He}}^{\text{+}}$, the total angular momentum $\left(J\right)$ for ${\text{He}}^{\text{+}}$ is given by the formula below.

  $J=\left|L-S\right|\mathrm{........................}\left(2\right)$

Substitute the value of $L$ and $S$ in equation (1).

  $\begin{array}{c}J=1-\frac{1}{2}\\ =\frac{1}{2}\end{array}$

Substitute the value of $S$, $J$ and the term for $L=1$ in equation (1).

Therefore the term symbol for the excited state of ${\text{He}}^{\text{+}}$ is shown below.

    ${}^{\text{2}}{\text{P}}_{\text{1}/\text{2}}$

The ground state configuration of ${\text{He}}^{\text{+}}$ is ${\text{1s}}^{\text{1}}$.

The value of $L$ is equal to the value of $l$.

The value of $l$ for the $\text{s}$ orbital is $0$.

Therefore the value of $L$ is $0$ which corresponds to the term $\text{S}$.

The value of $S$ for ${\text{He}}^{\text{+}}$ atom in the excited state is $\frac{1}{2}$

Therefore, the value of $2S+1$ is calculated below.

    $\begin{array}{c}2S+1=2\times \frac{1}{2}+1\\ =2\end{array}$

The total angular momentum $\left(J\right)$ for ${\text{He}}^{\text{+}}$ in the ground state is given by the formula below.

  $J=\left|L-S\right|\mathrm{........................}\left(2\right)$

Substitute the value of $L$ and $S$ in equation (1).

  $\begin{array}{c}J=\left|0-\frac{1}{2}\right|\\ =\frac{1}{2}\end{array}$

Substitute the value of $S$, $J$ and the term for $L=0$ in equation (1).

Therefore the term symbol for the excited state of ${\text{He}}^{\text{+}}$ is shown below.

    ${}^{\text{2}}{\text{S}}_{\text{1}/\text{2}}$

Therefore, the transition occurs from ${}^{\text{2}}{\text{S}}_{\text{1}/\text{2}}{\to }^{\text{2}}{\text{P}}_{\text{1}/\text{2}}$.

(b)

Interpretation Introduction

Interpretation:

The wavelength, frequency and wavenumber of the transition has to be calculated.

Concept introduction:

The wavelength, frequency and wavenumber of transition are all related to the energy of the transition.  The wavelength is inversely proportional to the energy of transition.  The frequency and wavenumber are directly proportional to energy of transition.

Answer to Problem 8.1IA

The frequency of the transition is $\underset{\_}{1.23\times 10^{16}{s}^{-1}}$.  The wavelength of the transition is $\underset{\_}{2.43\times 10^{-8}m}$.  The wavenumber of the transition is $\underset{\_}{0.411\times 10^8{m}^{-1}}$.

Explanation of Solution

The energy of transition of the hydrogen like atom is given below.

    $E_{n\to n\text{'}}=13.6Z^2\left(\frac{1}{n^2}-\frac{1}{{\left(n\text{'}\right)}^2}\right)\text{eV}\mathrm{..............}\left(3\right)$

Where,

$n$ and $n\text{'}$ is the number of the state.

$Z$ is the atomic number.

The atomic number of ${\text{He}}^{\text{+}}$ ($Z$) is $2$.

The transition takes place from $n=1$ to $n\text{'}=4$.

Substitute the value of $n$, and $n\text{'}$ in equation (3).

    $\begin{array}{c}{E}_{n\to n\text{'}}=13.6\times {\left(2\right)}^2\left(\frac{1}{1}-\frac{1}{{\left(4\right)}^2}\right)\text{eV}\\ =\text{54}\text{.4}\times \frac{15}{16}\text{eV}\\ =\text{51}\text{eV}\times \frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\\ =\text{8}{\text{.16×10}}^{\text{-18}}\text{J}\end{array}$

The formula to calculate frequency $\left(v\right)$ is given below.

    $E=hv\mathrm{...............}\left(4\right)$

Where,

$h$ is Planck’s constant.

The value of Planck’s constant ($h$) is $6.626\times {10}^{-34}\text{J}\text{s}$.

Substitute the value of $E$ in equation (4).

    $\begin{array}{c}8.16\times {10}^{-18}\text{J}=6.626\times {10}^{-34}\text{J}\text{s}\times v\\ v=\frac{8.16\times {10}^{-18}\text{J}}{6.626\times {10}^{-34}\text{J}\text{s}}\\ v=\underset{\_}{1.23\times 10^{16}{s}^{-1}}\end{array}$

Therefore, frequency of the transition is $\underset{\_}{1.23\times 10^{16}{s}^{-1}}$.

The wavelength ($\lambda$ ) is calculated by the formula below.

    $\lambda =\frac{c}{v}\mathrm{..............}\left(5\right)$

Where,

$c$ is the speed of light.

The speed of light is $3\times {10}^8\text{m}{\text{s}}^{-1}$.

Substitute the value of $v$ and $c$ in equation (5).

    $\begin{array}{c}\lambda =\frac{3\times {10}^8\text{m}{\text{s}}^{-1}}{\text{1}{\text{.23×10}}^{\text{16}}{\text{s}}^{-\text{1}}}\\ =\underset{\_}{2.43\times 10^{-8}m}\end{array}$

Therefore, wavelength of the transition is $\underset{\_}{2.43\times 10^{-8}m}$.

The relation wavelength $\left(\lambda \right)$ and wavenumber $\left(\tilde{\nu }\right)$ is given below.

    $\frac{1}{\lambda }=\tilde{\nu }\mathrm{..................}\left(6\right)$

Substitute the value of $\lambda$ in equation (2).

    $\begin{array}{c}\tilde{\nu }=\frac{1}{\text{2}{\text{.43×10}}^{-8}\text{m}}\\ =\underset{\_}{0.411\times 10^8{m}^{-1}}\end{array}$

Therefore, the wavenumber is $\underset{\_}{0.411\times 10^8{m}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The change in mean radius of electron during the transition has to be calculated.

Concept introduction:

The total distance from the nucleus of an atom to its outermost shell, where probability of finding the electron is maximum is termed as radius of the atom.  The most probable distance of the electron form nucleus in the ground state of hydrogen atom is termed as Bohr’s radius.

Answer to Problem 8.1IA

The change in mean radius during transition is $10.75a_0$.

Explanation of Solution

The mean radius of the electron is given by the formula shown below.

    $r_{n,l,m_l}=\frac{n^2{a}_0}{Z}\left\{1+\frac{1}{2}\left[1-\frac{l\left(l+1\right)}{n^2}\right]\right\}\mathrm{..................}\left(7\right)$

Where,

$Z$ is the atomic number.

$a_0$ is the Bohr’s radius.

The atomic number of ${\text{He}}^{\text{+}}$ ($Z$) is $2$.

Substitute the value of $Z$, $n=4$ and $l=1$ in equation (7).

    $\begin{array}{c}{r}_{n,l,m_l}=\frac{4^2{a}_0}{2}\left\{1+\frac{1}{2}\left[1-\frac{1\left(1+1\right)}{4^2}\right]\right\}\\ =8a_0\left\{1+\frac{1}{2}\left[1-\frac{1}{8}\right]\right\}\\ =8a_0\left\{\frac{23}{16}\right\}\\ =\frac{23a_0}{2}\end{array}$

The ground state configuration of ${\text{He}}^{\text{+}}$ is ${\text{1s}}^{\text{1}}$.

The value of $l$ for the $\text{s}$ orbital is $0$.

The value of $n$ is $1$.

Substitute the value of $Z$, $n=1$ and $l=0$ in equation (7).

    $\begin{array}{c}{r}_{n,l,m_l}=\frac{1^2{a}_0}{2}\left\{1+\frac{1}{2}\left[1-\frac{0\left(0+1\right)}{n^2}\right]\right\}\\ =\frac{1a_0}{2}\left\{1+\frac{1}{2}\right\}\\ =\frac{3a_0}{4}\end{array}$

Therefore, the change in radius during transition is calculated below.

    $\begin{array}{c}\text{Change in radius}=\frac{23a_0}{2}-\frac{3a_0}{4}\\ =\frac{43a_0}{4}\\ =10.75a_0\end{array}$

Therefore, the change in mean radius during transition is $10.75a_0$.