Chapter 27, Problem 44E

(a)

To determine

Energy emitted due to transition.

Answer to Problem 44E

The photon of energy emitted is $1081\text{eV}$.

Explanation of Solution

Write the expression for energy of diatomic molecule.

    $E_n=\frac{n\left(n+1\right){\hslash }^2}{2I}$        (I)

Here, $E_n$ is the energy of $n^{th}$ level, $n$ is the number of level, $\hslash$ is the reduced Planck’s constant and $I$ is the moment of inertia.

Write the expression for change in energy due to transition.

    $\Delta E=E_2-E_0$        (II)

Conclusion:

Substitute $2$ for $n$, $1.055\times {10}^{-34}\text{Js}$ for $\hslash$ and $1.92\times {10}^{-46}{\text{kgm}}^{\text{2}}$ for $I$ in equation (I).

    $\begin{array}{c}{E}_2=\frac{2\left(2+1\right){\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{2\left(1.92\times {10}^{-46}{\text{kgm}}^2\right)}\\ =\frac{2\left(3\right){\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{2\left(1.92\times {10}^{-46}{\text{kgm}}^2\right)}\left(\frac{1{\text{kgm}}^2{\text{s}}^{-2}}{1\text{J}}\right)\\ =1.73\times {10}^{-22}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =1081\text{eV}\end{array}$

 Substitute $0$ for $n$, $1.055\times {10}^{-34}\text{Js}$ for $\hslash$ and $1.92\times {10}^{-46}{\text{kgm}}^{\text{2}}$ for $I$ in equation (I).

    $\begin{array}{c}{E}_0=\frac{\left(0\right)\left(0+1\right){\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{2\left(1.92\times {10}^{-46}{\text{kgm}}^2\right)}\\ =0\end{array}$

Substitute $1081\text{eV}$ for $E_2$ and $0\text{eV}$ for $E_0$ in equation (II)>

    $\begin{array}{c}\Delta E=1081\text{eV}-0\text{eV}\\ \text{=}1081\text{eV}\end{array}$

Thus, the energy of photon emitted is $1081\text{eV}$.

(b)

To determine

The wavelength of photon emitted.

Answer to Problem 44E

The wavelength of photon emitted is $11.49\text{nm}$.

Explanation of Solution

Write the expression for energy of photon emitted.

    $E=\frac{hc}{\lambda }$

Here, $E$ is the energy of photon, $h$ is Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength.

Rearrange the above expression for $\lambda$.

    $\lambda =\frac{hc}{E}$        (III)

Conclusion:

Substitute $6.626\times {10}^{-34}\text{Js}$ for $h$, $3\times {10}^8{\text{ms}}^{-1}$ for $c$ and $1081\text{eV}$ for $E$ in equation (III).

    $\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\text{Js}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{1081\text{eV}}\\ =\frac{\left(6.626\times {10}^{-34}\text{Js}\right)\left(3\times {10}^8{\text{ms}}^{-1}\right)}{1081\text{eV}}\left(\frac{1{\text{kgm}}^2{\text{s}}^{-2}}{1\text{J}}\right)\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =1.149\times {10}^{-8}\text{m}\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =11.49\text{nm}\end{array}$

Thus, the wavelength of photon emitted is $11.49\text{nm}$.