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Chapter 7: Problem 63

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=3 \rightarrow n=2\) b. \(n=4 \rightarrow n=2\) c. \(n=2 \rightarrow n=1\)

Short Answer

Expert verified
For the given electronic transitions in the hydrogen atom: a. \(n=3 \rightarrow n=2\): The wavelength of light emitted is approximately \(6.56 \times 10^{-7}\, \text{m}\), which is visible red light. b. \(n=4 \rightarrow n=2\): The wavelength of light emitted is approximately \(4.86 \times 10^{-7}\, \text{m}\), which is visible blue-green light. c. \(n=2 \rightarrow n=1\): The wavelength of light emitted is approximately \(1.22 \times 10^{-7}\, \text{m}\), which is ultraviolet radiation.

Step by step solution

01

Understand the Rydberg Formula

The Rydberg formula is used to calculate the wavelength of light emitted or absorbed during electronic transitions in atoms. For the hydrogen atom, the simplified Rydberg formula is: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] where: - \(\lambda\) is the wavelength of light - \(R_H\) is the Rydberg constant for hydrogen, approximately equal to \(1.097 \times 10^7 \, \text{m}^{-1}\) - \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final energy levels, respectively (with \(n_1 < n_2\))
02

Use the Rydberg Formula to Calculate the Wavelengths

For each of the transitions provided, plug in the corresponding values of \(n_1\) and \(n_2\) into the Rydberg formula, and solve for \(\lambda\). a. \(n=3 \rightarrow n=2\) \[\frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2}\right)\] b. \(n=4 \rightarrow n=2\) \[\frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{4^2}\right)\] c. \(n=2 \rightarrow n=1\) \[\frac{1}{\lambda} = R_H \left(\frac{1}{1^2} - \frac{1}{2^2}\right)\]
03

Evaluate Each Expression

Evaluate each expression for \(\lambda\) using the given value of \(R_H\). a. \(n=3 \rightarrow n=2\) \[\lambda = \frac{1}{R_H \left(\frac{1}{2^2} - \frac{1}{3^2}\right)} \approx 6.56 \times 10^{-7} \, \text{m}\] b. \(n=4 \rightarrow n=2\) \[\lambda = \frac{1}{R_H \left(\frac{1}{2^2} - \frac{1}{4^2}\right)} \approx 4.86 \times 10^{-7} \, \text{m}\] c. \(n=2 \rightarrow n=1\) \[\lambda = \frac{1}{R_H \left(\frac{1}{1^2} - \frac{1}{2^2}\right)} \approx 1.22 \times 10^{-7} \, \text{m}\]
04

Determine the Type of Electromagnetic Radiation

Compare each wavelength to the electromagnetic spectrum to classify the type of radiation emitted in each transition. a. \(n=3 \rightarrow n=2\): \(\lambda \approx 6.56 \times 10^{-7}\, \text{m}\) corresponds to visible light (specifically, red light). b. \(n=4 \rightarrow n=2\): \(\lambda \approx 4.86 \times 10^{-7}\, \text{m}\) corresponds to visible light (specifically, blue-green light). c. \(n=2 \rightarrow n=1\): \(\lambda \approx 1.22 \times 10^{-7}\, \text{m}\) corresponds to ultraviolet radiation. Thus, the electronic transitions emit visible light for (a) and (b), and ultraviolet radiation for (c).

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Most popular questions from this chapter

One of the visible lines in the hydrogen emission spectrum corresponds to the \(n=6\) to \(n=2\) electronic transition. What color light is this transition? See Exercise 150 .

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