- #1
EliotHijano
- 18
- 0
Hello,
I would like to know how to calculate the broadening of the spectral lines caused by the Doppler effect for the Lyman, Balmer and Paschen series. To be more concrete, I would like to know the broadening of the alpha transitions.
The equations I use are the following but i don't know if I am doing something wrong.
[tex]\Delta \nu &=&2\frac{\nu _{o}}{c}\sqrt{\frac{2KT}{m}\ln \left( 2\right) }[/tex]
I calculate [tex]\nu _{o}[/tex] Doing the following:
[tex]E_{n}-E_{n^{\prime }} &=&\left[ \frac{1}{\left( n^{\prime }\right) ^{2}}-
\frac{1}{\left( n\right) ^{2}}\right] \frac{Z^{2}e^{4}\mu }{2\left( 4\pi
\varepsilon _{o}\right) ^{2}\hbar ^{2}}=-h\nu _{o}[/tex]
[tex]\nu _{o} &=&-\frac{Z^{2}e^{4}\mu }{4\pi \left( 4\pi \varepsilon _{o}\right)
^{2}\hbar ^{3}}\left[ \frac{1}{\left( n^{\prime }\right) ^{2}}-\frac{1}{
\left( n\right) ^{2}}\right][/tex]
For T=300K we have:
[tex]\nu _{o} &\approx &-\frac{e^{4}m_{e}}{4\pi \left( 4\pi \varepsilon
_{o}\right) ^{2}\hbar ^{3}}\left[ \frac{1}{\left( n^{\prime }\right) ^{2}}-
\frac{1}{\left( n\right) ^{2}}\right] \approx -3.288953357\cdot 10^{15}\left[
\frac{1}{\left( n^{\prime }\right) ^{2}}-\frac{1}{\left( n\right) ^{2}}
\right] \ \ Hz[/tex]
SO:
[tex]\Delta \nu &=&2\frac{\nu _{o}}{c}\sqrt{\frac{2KT}{m}\ln \left( 2\right) }
\approx -0.000040625\cdot 10^{15}\left[ \frac{1}{\left( n^{\prime }\right)
^{2}}-\frac{1}{\left( n\right) ^{2}}\right] \ \ Hz[/tex]
Finally, the following numbers are obtained:
[tex]\begin{tabular}{|l|l|}
\hline
$Line$ & $\Delta \nu \ (GHz)$ \\ \hline\hline
$\alpha \ LYMAN$ & $ 30.4685$ \\ \hline
$\alpha \ BALMER$ & $ 5.64236$ \\ \hline
$\alpha \ PASCHEN$ & $ 1.974826$ \\ \hline
\end{tabular}[/tex]
Unfortunatelly, I can't find any book to confirm this results, that is why I am posting this.
What do you say? Am I doing anything wrong?
Eliot.
I would like to know how to calculate the broadening of the spectral lines caused by the Doppler effect for the Lyman, Balmer and Paschen series. To be more concrete, I would like to know the broadening of the alpha transitions.
The equations I use are the following but i don't know if I am doing something wrong.
[tex]\Delta \nu &=&2\frac{\nu _{o}}{c}\sqrt{\frac{2KT}{m}\ln \left( 2\right) }[/tex]
I calculate [tex]\nu _{o}[/tex] Doing the following:
[tex]E_{n}-E_{n^{\prime }} &=&\left[ \frac{1}{\left( n^{\prime }\right) ^{2}}-
\frac{1}{\left( n\right) ^{2}}\right] \frac{Z^{2}e^{4}\mu }{2\left( 4\pi
\varepsilon _{o}\right) ^{2}\hbar ^{2}}=-h\nu _{o}[/tex]
[tex]\nu _{o} &=&-\frac{Z^{2}e^{4}\mu }{4\pi \left( 4\pi \varepsilon _{o}\right)
^{2}\hbar ^{3}}\left[ \frac{1}{\left( n^{\prime }\right) ^{2}}-\frac{1}{
\left( n\right) ^{2}}\right][/tex]
For T=300K we have:
[tex]\nu _{o} &\approx &-\frac{e^{4}m_{e}}{4\pi \left( 4\pi \varepsilon
_{o}\right) ^{2}\hbar ^{3}}\left[ \frac{1}{\left( n^{\prime }\right) ^{2}}-
\frac{1}{\left( n\right) ^{2}}\right] \approx -3.288953357\cdot 10^{15}\left[
\frac{1}{\left( n^{\prime }\right) ^{2}}-\frac{1}{\left( n\right) ^{2}}
\right] \ \ Hz[/tex]
SO:
[tex]\Delta \nu &=&2\frac{\nu _{o}}{c}\sqrt{\frac{2KT}{m}\ln \left( 2\right) }
\approx -0.000040625\cdot 10^{15}\left[ \frac{1}{\left( n^{\prime }\right)
^{2}}-\frac{1}{\left( n\right) ^{2}}\right] \ \ Hz[/tex]
Finally, the following numbers are obtained:
[tex]\begin{tabular}{|l|l|}
\hline
$Line$ & $\Delta \nu \ (GHz)$ \\ \hline\hline
$\alpha \ LYMAN$ & $ 30.4685$ \\ \hline
$\alpha \ BALMER$ & $ 5.64236$ \\ \hline
$\alpha \ PASCHEN$ & $ 1.974826$ \\ \hline
\end{tabular}[/tex]
Unfortunatelly, I can't find any book to confirm this results, that is why I am posting this.
What do you say? Am I doing anything wrong?
Eliot.