The wavelengths of the Paschen series for hydrogen are given...

The wavelengths of the Paschen series for hydrogen are given by $$\frac{1}{\lambda}=R_{\mathrm{H}}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) \quad n=4,5,6, \ldots$$ (a) Calculate the wavelengths of the first three lines in this series. (b) Identify the region of the electromagnetic spectrum in which these lines appear.

The wavelengths of the Paschen series for hydrogen are given by
$$\frac{1}{\lambda}=R_{\mathrm{H}}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) \quad n=4,5,6, \ldots$$
(a) Calculate the wavelengths of the first three lines in this series. (b) Identify the region of the electromagnetic spectrum in which these lines appear.

Key Concepts

Hydrogen Atomic Spectrum

This concept involves understanding that hydrogen emits or absorbs light at specific wavelengths when its electron transitions between defined energy levels. The pattern of these emissions is not continuous but appears as discrete lines, each corresponding to a specific quantum jump. These spectral lines form characteristic series named after the lowest energy level involved in the electron transition, such as the Lyman, Balmer, or Paschen series.

Paschen Series

The Paschen series is one of the spectral series of the hydrogen atom where the electron transitions end at the n=3 energy level. When electrons drop from higher levels (n > 3) to n=3, they emit light whose wavelengths fall in the infrared region of the electromagnetic spectrum. This series is significant in the study of quantum mechanics and spectroscopy, specifically in infrared applications.

Rydberg Formula

The Rydberg formula is a mathematical expression used to calculate the wavelengths of spectral lines emitted or absorbed by hydrogen and hydrogen-like atoms. It relates the inverse wavelength to the difference between the inverse squares of the initial and final principal quantum numbers, multiplied by the Rydberg constant. This formula underpins the prediction and explanation of the spectral series observed in hydrogen.

Electromagnetic Spectrum

This refers to the entire range of electromagnetic radiation, from gamma rays with very short wavelengths to radio waves with very long wavelengths. Understanding where a particular spectral line falls within the electromagnetic spectrum is important for analyzing the properties and potential applications of the radiation. For example, the infrared region, where the Paschen series lies, has wavelengths longer than visible light and is important in applications like thermal imaging and spectroscopy.

Transcript

00:01 Calculate the first three wavelengths of the passion series.

00:06 So the equation that governs the passion series is 1 over lambda is equal to the rydberg constant times 1 over 3 squared minus 1 over n squared.

00:14 So the first three wavelengths are going to be n equals 4, n equals 5, and n equals 6.

00:19 So i flipped the equation.

00:22 So we want to solve for lambda.

00:23 So lambda is equal to the inverse of the value that's on the right side of the equation.

00:27 So i took everything on the right side of the equation and raised it to the negative 1.

00:31 So now let's start with the first wavelength.

00:33 So this is going to be 4n is equal to 4.

00:37 So the transition from 4 to 3.

00:39 So we can call this lambda 1 because it's the first wavelength in this series.

00:43 And this is just going to be equal to the redbird constant multiplied by we have 1 over 3 squared, which is 9, minus 1 over 4 squared, which is 16.

01:07 Of course, all of this is going to get raised to the negative 1.

01:19 The riddberg constant here is 1 .097 times 10 to the 7 inverse meters.

01:27 So plugging that value in and carrying out this operation, we find that the first wavelength is equal to 1 ,875 0 .875 nanometers.

01:43 We can box that in as a solution for the first part of the question.

01:47 So now we're just going to do the exact same thing.

01:49 For n equals 5 or n equals 5 and n equals 6.

01:55 So n equals 5, we have lambda 2...

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