Chapter 8, Problem 8C.3P

Interpretation Introduction

Interpretation:

The presence of both ${}^4{\text{He}}^{+}$ and ${}^3{\text{He}}^{+}$ in a star with use of spectroscopycan be confirmed by calculation of wavenumber for $n=3\to n=2$ and $n=2\to n=1$ transitions for each ionic isotope have to be shown.

Concept introduction:

An atom is made up of three subatomic particles-neutrons, protons, and electrons.  Neutron and protons are present in the nucleus of the atom, whereas electrons are revolving outside the nucleus in an atom.  An electron can excite or relax from one energy level to another energy level.

Answer to Problem 8C.3P

When the spectroscopy data matches with the data in the table shown below, then it confirms that the ${}^4{\text{He}}^{+}$ and ${}^3{\text{He}}^{+}$ ions are present in star.

Transition	Ion	Wavenumber
$n=3\to n=2$	${}^3{\text{He}}^{+}$	$60959.5574{\text{cm}}^{-1}$
$n=3\to n=2$	${}^4{\text{He}}^{+}$	$60962.3016{\text{cm}}^{-1}$
$n=2\to n=1$	${}^3{\text{He}}^{+}$	$329155.2777{\text{cm}}^{-1}$
$n=2\to n=1$	${}^4{\text{He}}^{+}$	$329170.0952{\text{cm}}^{-1}$

Explanation of Solution

The mass of ${}^4{\text{He}}^{+}$ nucleus is $4.00260\text{u}$.

The mass of ${}^4{\text{He}}^{+}$ nucleus in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{n}}\left({}^4{\text{He}}^{+}\right)=\left(4.00260\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =6.64648\times {10}^{-27}\text{kg}\end{array}$

The mass of ${}^3{\text{He}}^{+}$ nucleus is $3.01603\text{u}$.

The mass of ${}^3{\text{He}}^{+}$ nucleus in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{n}}\left({}^3{\text{He}}^{+}\right)=\left(3.01603\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =5.00824\times {10}^{-27}\text{kg}\end{array}$

The reduced mass of an atom is given by the expression as shown below.

  $\mu =\frac{m_{\text{e}}{m}_{\text{n}}}{m_{\text{e}}+m_{\text{n}}}$                                                                                                   (1)

Where,

• $m_{\text{e}}$ is the mass of an electron with a value of $9.10938\times {10}^{-31}\text{kg}$.
• $m_{\text{n}}$ is the mass of the nucleus.

Substitute the values of $m_{\text{e}}$ and mass of ${}^4{\text{He}}^{+}$ nucleus in the equation (1).

  $\begin{array}{c}\mu =\frac{\left(9.10938\times {10}^{-31}\text{kg}\right)\left(6.64648\times {10}^{-27}\text{kg}\right)}{\left(9.10938\times {10}^{-31}\text{kg}\right)+\left(6.64648\times {10}^{-27}\text{kg}\right)}\\ =9.10813\times {10}^{-31}\text{kg}\end{array}$

Substitute the values of $m_{\text{e}}$ and mass of ${}^3{\text{He}}^{+}$ nucleus in the equation (1).

  $\begin{array}{c}\mu =\frac{\left(9.10938\times {10}^{-31}\text{kg}\right)\left(5.00824\times {10}^{-27}\text{kg}\right)}{\left(9.10938\times {10}^{-31}\text{kg}\right)+\left(5.00824\times {10}^{-27}\text{kg}\right)}\\ =9.10772\times {10}^{-31}\text{kg}\end{array}$

The wavenumber of a one electron ion is given by the expression shown below.

  $\overline{\nu }=\frac{\mu e^4{Z}^2}{8{\epsilon }_0^2{h}^{3}c}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$                                                                                    (2)

Where,

• $Z$ is the atomic number of atom.
• ${\epsilon }_0$ is the permittivity of the vacuum.
• $h$ is Plank’s constant with a value of $6.62608\times {10}^{-34}\text{Js}$.
• $c$ is the speed of light with value of $2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}$.
• $\mu$ is the reduced mass of atom.
• $n_{\text{1}}$ is the lower energy level.
• $n_{\text{2}}$ is the higher energy level.
• $e$ is charge on electron.

The value of $Z$ for helium atom is $2$.

The charge on electron is $1.60218\times {10}^{-19}\text{C}$.

The permittivity of the vacuum is $8.85419\times {10}^{-14}{\text{J}}^{-1}{\text{C}}^2{\text{cm}}^{-1}$.

Substitute the value of $Z$, $e$, $c$, $h$, and ${\epsilon }_0$ in equation (2).

  $\begin{array}{c}\overline{\nu }=\left(\begin{array}{c}\frac{\mu {\left(1.60218\times {10}^{-19}\text{C}\right)}^4{\left(2\right)}^2}{8{\left(8.85419\times {10}^{-14}{\text{J}}^{-1}{\text{C}}^2{\text{cm}}^{-1}\right)}^2{\left(6.62608\times {10}^{-34}\text{Js}\right)}^3\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)}\\ \times \left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)\end{array}\right)\\ \overline{\nu }=\frac{2.6358\times {10}^{-75}}{5.46992\times {10}^{-115}}\left(\mu \right)\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)\left({\text{Js}}^2{\text{cm}}^{-1}\right)\left(\frac{10000\text{kg}{\text{cm}}^2{\text{s}}^{-2}}{1\text{J}}\right)\\ \overline{\nu }=\left(4.8187\times {10}^{35}\right)\left(\mu \right)\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right){\text{kg}}^{-1}{\text{cm}}^{-1}\end{array}$

Therefore, the expression of wavenumber for helium atom is shown below.

  $\overline{\nu }=\left(4.8187\times {10}^{35}\right)\left(\mu \right)\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right){\text{kg}}^{-1}{\text{cm}}^{-1}$                                                 (3)

Substitute the value of reduced mass of ${}^4{\text{He}}^{+}$ , $n_{\text{1}}=2$ and $n_{\text{2}}=1$ in the equation (3).

    $\begin{array}{c}\overline{\nu }=\left(4.8187\times {10}^{35}\right)\left(9.10813\times {10}^{-31}\text{kg}\right)\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right){\text{kg}}^{-1}{\text{cm}}^{-1}\\ =\left(438893.4603\right)\left(0.75\right)\\ =329170.0952{\text{cm}}^{-1}\end{array}$

Substitute the value of reduced mass of ${}^4{\text{He}}^{+}$ , $n_{\text{1}}=3$ and $n_{\text{2}}=2$ in the equation (3).

    $\begin{array}{c}\overline{\nu }=\left(4.81870\times {10}^{35}\right)\left(9.10813\times {10}^{-31}\text{kg}\right)\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}\right){\text{kg}}^{-1}{\text{cm}}^{-1}\\ =\left(438893.4603\right)\left(0.1389\right)\\ =60962.3016{\text{cm}}^{-1}\end{array}$

Substitute the value of reduced mass of ${}^3{\text{He}}^{+}$ , $n_{\text{1}}=2$ and $n_{\text{2}}=1$ in the equation (3).

    $\begin{array}{c}\overline{\nu }=\left(4.8187\times {10}^{35}\right)\left(9.10772\times {10}^{-31}\text{kg}\right)\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right){\text{kg}}^{-1}{\text{cm}}^{-1}\\ =\left(438873.7036\right)\left(0.75\right){\text{cm}}^{-1}\\ =329155.2777{\text{cm}}^{-1}\end{array}$

Substitute the value of reduced mass of ${}^3{\text{He}}^{+}$ , $n_{\text{1}}=3$ and $n_{\text{2}}=2$ in the equation (3).

    $\begin{array}{c}\overline{\nu }=\left(4.81870\times {10}^{35}\right)\left(9.10772\times {10}^{-31}\text{kg}\right)\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}\right){\text{kg}}^{-1}{\text{cm}}^{-1}\\ =\left(438873.7036\right)\left(0.1389\right)\\ =60959.5574{\text{cm}}^{-1}\end{array}$

The wavenumber of ${}^3{\text{He}}^{+}$ and ${}^4{\text{He}}^{+}$ for $n=3\to n=2$ and $n=2\to n=1$ transitions are summarized in the table shown below.

Transition	Ion	Wavenumber
$n=3\to n=2$	${}^3{\text{He}}^{+}$	$60959.5574{\text{cm}}^{-1}$
$n=3\to n=2$	${}^4{\text{He}}^{+}$	$60962.3016{\text{cm}}^{-1}$
$n=2\to n=1$	${}^3{\text{He}}^{+}$	$329155.2777{\text{cm}}^{-1}$
$n=2\to n=1$	${}^4{\text{He}}^{+}$	$329170.0952{\text{cm}}^{-1}$

When the spectroscopy data matches with the data in the above table, then it can confirm that ${}^4{\text{He}}^{+}$ and ${}^3{\text{He}}^{+}$ ions are present in star.

Therefore, the presence of both ${}^4{\text{He}}^{+}$ and ${}^3{\text{He}}^{+}$ in a star with use of spectroscopy can be confirmed by calculation of wavenumber for $n=3\to n=2$ and $n=2\to n=1$ transitions for each ionic isotope.