Chapter 41, Problem 13P

(a)

To determine

To show that the difference in wavelength between the hydrogen-1 and deuterium spectral lines associated with a particular electron transition is ${\lambda }_{\text{H}}-{\lambda }_{\text{D}}=\left(1-\frac{{\mu }_{\text{H}}}{{\mu }_{\text{D}}}\right){\lambda }_{\text{H}}$.

Answer to Problem 13P

The difference in wavelength between the hydrogen-1 and deuterium spectral lines associated with a particular electron transition is ${\lambda }_{\text{H}}-{\lambda }_{\text{D}}=\left(1-\frac{{\mu }_{\text{H}}}{{\mu }_{\text{D}}}\right){\lambda }_{\text{H}}$.

Explanation of Solution

Write the general expression for the energy levels of one electron atoms.

    $E_n=-\frac{\mu k_e{}^2{q}_1^2{q}_2^2}{2{\hslash }^2{n}^2}$

Here, $E_n$ is the energy, $\mu$ is the reduced mass of the atom, $\hslash$ is the Planck’s constant, $n$ is the principle quantum number, $k_e$ is the Coulomb’s constant, $q_1$ is the charge of electron, and $q_2$ is the charge of the nucleus.

Refer the above equation and write an expression for energy released for any transition in hydrogen-1 or proton atom.

    $\Delta E_P=-\frac{{\mu }_P{k}_e{}^2{e}^4}{2{\hslash }^2}\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)$                                                                               (I)

Here, $\Delta E_P$ is the energy of the transition of proton, $e$ is the charge of an electron, ${\mu }_P$ is the reduced mass of proton atom, $n_i$ is the principle quantum number of the initial state, and $n_f$ is the principle quantum number of the final state.

Refer the above equation and write an expression for energy released for any transition in hydrogen-2 or deuterium atom.

    $\Delta E_D=-\frac{{\mu }_D{k}_e{}^2{e}^4}{2{\hslash }^2}\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)$                                                                               (II)

Here, $\Delta E_D$ is the energy of transition of deuterium, and ${\mu }_D$ is the reduced mass of deuterium.

Write the formula for the reduced mass of proton atom.

    ${\mu }_P=\frac{m_e{m}_P}{m_e+m_P}$                                                                                                    (III)

Here, $m_e$ is the mass of electron, and $m_P$ is the mass of proton nucleus.

Write the formula for the reduced mass of the deuterium atom.

    ${\mu }_{\text{D}}=\frac{m_e{m}_{\text{D}}}{m_e+m_{\text{D}}}$                                                                                                   (IV)

Here, $m_D$ is the mass of deuterium nucleus.

Divide equation (I) and (II).

    $\frac{\Delta E_{\text{H}}}{\Delta E_{\text{D}}}=\frac{{\mu }_{\text{H}}}{{\mu }_{\text{D}}}$

Energy is inversely proportional to the wavelength. Thus re-write the above equation.

    $\begin{array}{c}\frac{{\lambda }_D}{{\lambda }_P}=\frac{{\mu }_P}{{\mu }_D}\\ {\lambda }_D=\frac{{\mu }_P}{{\mu }_D}{\lambda }_P\end{array}$

Here, ${\lambda }_D$ is the wave length of the spectral line of the deuterium atom, and ${\lambda }_P$ is the wavelength of the spectral line of the proton atom.

Re-write the above equation.

    ${\lambda }_{\text{H}}-{\lambda }_{\text{D}}=\left(1-\frac{{\mu }_{\text{H}}}{{\mu }_{\text{D}}}\right){\lambda }_{\text{H}}$

Conclusion:

The difference in wavelength between the hydrogen-1 and deuterium spectral lines associated with a particular electron transition is ${\lambda }_{\text{H}}-{\lambda }_{\text{D}}=\left(1-\frac{{\mu }_{\text{H}}}{{\mu }_{\text{D}}}\right){\lambda }_{\text{H}}$.

(b)

To determine

The wavelength difference for the Balmer alpha line of hydrogen emitted from a transition from $n=3$ to $n=4$.

Answer to Problem 13P

The wavelength difference for the Balmer alpha line of hydrogen emitted from a transition from $n=3$ to $n=4$ is $0.179\text{nm}$.

Explanation of Solution

Refer section (a) and write the formula for the ration of the reduced mass of proton.

    ${\mu }_H=\frac{m_e{m}_P}{m_e+m_P}$                                                                                                    (III)

Here, ${\mu }_H$ is the reduced mass of hydrogen 1, $m_e$ is the mass of electron, and $m_P$ is the mass of proton nucleus.

Write the formula for the reduced mass of the deuterium atom.

    ${\mu }_{\text{D}}=\frac{m_e{m}_{\text{D}}}{m_e+m_{\text{D}}}$                                                                                                   (IV)

Here, ${\mu }_{\text{D}}$ is the reduced mass of deuterium, and $m_D$ is the mass of deuterium nucleus.

Write the formula for the ration of reduced mass of proton and deuterium.

    $\frac{{\mu }_{\text{H}}}{{\mu }_{\text{D}}}=\left(\frac{m_e{m}_p}{m_e+m_p}\right)\left(\frac{m_e+m_{\text{D}}}{m_e{m}_{\text{D}}}\right)$                                                                           (V)

Write the formula for the difference in wavelength of proton and deuterium.

    ${\lambda }_{\text{H}}-{\lambda }_{\text{D}}=\left(1-\frac{{\mu }_{\text{H}}}{{\mu }_{\text{D}}}\right){\lambda }_{\text{H}}$                                                                                    (VI)

Conclusion:

Substitute $1.007276\text{u}$ for $m_p$, $0.000549\text{u}$ for $m_e$, $2.013553\text{u}$ for $m_D$ in equation (V).

    $\begin{array}{c}\frac{{\mu }_{\text{H}}}{{\mu }_{\text{D}}}=\frac{\left(1.007276\text{u}\right)\left(0.000549\text{u}+2.013553\text{u}\right)}{\left(0.000549\text{u}+1.007276\text{u}\right)\left(2.013553\text{u}\right)}\\ =0.999728\end{array}$

Substitute $0.999728$ for ${\mu }_H/{\mu }_D$, $656.3\text{nm}$ for ${\lambda }_H$ in equation (VI).

    $\begin{array}{c}{\lambda }_{\text{H}}-{\lambda }_{\text{D}}=\left(1-0.999728\right)\left(656.3\text{nm}\right)\\ =0.179\text{nm}\end{array}$

The wavelength difference for the Balmer alpha line of hydrogen emitted from a transition from $n=3$ to $n=4$ is $0.179\text{nm}$.