Using the rydberg constant determined in question 1, 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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