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Chapter 6: Problem 14

The Brackett series lines in the atomic spectrum of hydrogen result from transitions from \(\mathbf{n}>4\) to \(\mathbf{n}=4\) (a) Calculate the wavelength, in nanometers, of a line in this series resulting from the \(\mathbf{n}=5\) to \(\mathbf{n}=4\) transition. (b) In what region of the spectrum are these lines formed?

Short Answer

Expert verified
Question: Calculate the wavelength corresponding to the transition from n=5 to n=4 in the Brackett series of the hydrogen atomic spectrum, and determine which region of the spectrum these lines are formed in. Answer: The wavelength corresponding to the transition from n=5 to n=4 in the Brackett series is 1940 nm. The Brackett series lines are formed in the infrared region of the spectrum.

Step by step solution

01

Using the Rydberg formula for hydrogen's atomic spectrum

The Rydberg formula for hydrogen's atomic spectrum is given by: $$\frac{1}{\lambda}=R_H\left(\frac{1}{n^2_1} - \frac{1}{n^2_2}\right)$$ Where \(\lambda\) is the wavelength, \(n_1\) and \(n_2\) are the initial and final energy levels, and \(R_H\) is the Rydberg constant for hydrogen which is equal to \(1.097\times10^7 m^{-1}\). For this problem, we are given the transition from \(n=5\) to \(n=4\). So, \(n_1=4\) and \(n_2=5\).
02

Calculate the reciprocal of the wavelength

Plugging the values into the Rydberg formula, we get: $$\frac{1}{\lambda}=R_H\left(\frac{1}{4^2} - \frac{1}{5^2}\right)$$ Replacing the Rydberg constant: $$\frac{1}{\lambda} = (1.097\times10^7) \left(\frac{1}{16}-\frac{1}{25}\right)$$
03

Calculate the wavelength

To find the wavelength in meters, we must take the reciprocal of the value calculated in step 2: $$\lambda = \frac{1}{(1.097\times10^7) (\frac{1}{16}-\frac{1}{25})}$$ Calculating this result, we obtain the wavelength in meters: $$\lambda = 1.94\times10^{-6}\,m$$
04

Convert wavelength to nanometers

Now we must convert the wavelength from meters to nanometers. There are \(10^9\) nanometers in a meter, so we can multiply the result by this factor: $$\lambda_{nm} = \lambda \times 10^9 = (1.94\times10^{-6})(10^9)$$ This gives us the wavelength in nanometers: $$\lambda = 1940\, nm$$
05

Solution for part (a)

The wavelength corresponding to the transition from \(n=5\) to \(n=4\) in the Brackett series is \(1940\,nm\).
06

Identify the region of the spectrum

The electromagnetic spectrum has different regions based on wavelength. To determine in which region the Brackett series lines are formed, we must compare the wavelength calculated in part (a) to those regions: - Ultraviolet: \(10-400\,nm\) - Visible light: \(400-700\,nm\) - Infrared: \(700\,nm - 1\,mm\) With a wavelength of \(1940\,nm,\) the lines corresponding to the Brackett series fall in the infrared region of the spectrum.
07

Solution for part (b)

The Brackett series lines are formed in the infrared region of the spectrum.

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