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The shortest wavelength in the paschen series= [tex]8.2\times 10^{-7}[/tex] m.
How do we calculate the shortest wavelength in the paschen series?
Emission lines for hydrogen occur when electrons drop from some energy level to a lower energy level. To calculate the shortest wavelength in the paschen series we are using the formula,
[tex]\frac{1}{\lambda} =R_{H} (\frac{1}{n_{f}^{2} }-\frac{1}{n^{2} } )[/tex]
Here, we are given,
[tex]R_{H}[/tex]= Rydberg constant=[tex]1.09737 \times 10^{7} m^{-1}[/tex]
[tex]n_f[/tex]= The lower energy level quantum number.=3 (for the paschen series).
n= The quantum number of whichever state the transitions occur from = (for this case of the paschen series).
We have to find the wavelength associated with the photon emitted = [tex]\lambda[/tex] m.
Now we substitute the known values in the above equation, we can find that,
[tex]\frac{1}{\lambda} =1.09737 \times 10^{7} (\frac{1}{3^2 }-\frac{1}{\infty^{2} } )[/tex]
Or,[tex]\frac{1}{\lambda} =1.09737 \times 10^{7}\times \frac{1}{9 }[/tex]
Or,[tex]\frac{1}{\lambda} =1,219,300[/tex]
Or,[tex]\lambda= 8.2\times 10^{-7}[/tex] m
From the above calculation we can conclude that the shortest wavelength in the paschen series is [tex]8.2\times 10^{-7}[/tex] m
Learn more about paschen series:
https://brainly.com/question/15322810
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