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Chapter 6: Problem 92

Determine the wavelength of the third Balmer line (transition from \(n=5\) to \(n=2\) ).

Short Answer

Expert verified
The wavelength of the third Balmer line, corresponding to a transition from \(n=5\) to \(n=2\), is approximately \(4.43 \times 10^{-7}\, \text{m}\) or \(443\, \text{nm}\).

Step by step solution

01

Identify the initial and final energy levels

In this exercise, we are asked to find the wavelength of the third Balmer line, which corresponds to a transition from \(n=5\) to \(n=2\). Thus, our initial energy level (\(n_1\)) is 5, and our final energy level (\(n_2\)) is 2.
02

Apply the Balmer formula

The Balmer formula relates the wavelength of a photon emitted during the transition of an electron in a hydrogen atom to the initial and final energy levels: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right)\] where \(\lambda\) is the wavelength of the photon, \(R_H\) is the Rydberg constant for hydrogen (approximately \(1.097 \times 10^7\, \text{m}^{-1}\)), \(n_1\) is the initial energy level and \(n_2\) is the final energy level.
03

Plug in the values of the energy levels into the Balmer formula

Now, plug in the values of \(n_1 = 5\) and \(n_2 = 2\) and the Rydberg constant \(R_H\) into the Balmer formula: \[\frac{1}{\lambda} = 1.097 \times 10^7\, \text{m}^{-1} \left(\frac{1}{2^2} - \frac{1}{5^2}\right)\]
04

Calculate the reciprocal of the wavelength

Carry out the arithmetic inside the parenthesis: \[\frac{1}{\lambda} = 1.097 \times 10^7\, \text{m}^{-1} \times \left(\frac{1}{4} - \frac{1}{25}\right) = 1.097 \times 10^7\, \text{m}^{-1} \times \frac{21}{100}\]
05

Find the wavelength

Now, divide the Rydberg constant by the result obtained in step 4 to find the wavelength: \[\lambda = \frac{1}{1.097 \times 10^7\, \text{m}^{-1} \times \frac{21}{100}}\] Calculate the wavelength: \[\lambda = \frac{100}{1.097 \times 10^7\, \text{m}^{-1} \times 21}\] \[\lambda \approx 4.43 \times 10^{-7}\, \text{m}\] So, the wavelength of the third Balmer line is approximately \(4.43 \times 10^{-7}\, \text{m}\) or \(443\, \text{nm}\).

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