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The wavelength of the third line in the Lyman series, and identify the type of EM radiation

In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.

1 / lambda = R(h)* ( [tex]\frac{1}{(n1)^{2} }[/tex] -   [tex]\frac{1}{(n2)^{2} }[/tex])

                 = 109678 ( [tex]\frac{1}{1^{2} }[/tex] -  [tex]\frac{1}{3^{2} }[/tex] )

                 = 109678 (8/9)

   Lambda = 9 / (109678 * 8 )

                  = 102.6 * [tex]10^{-9}[/tex] m = 102.6 nm

To learn more about Lyman series here

https://brainly.com/question/5762197

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