Ch.6 - Electronic Structure of Atoms

Brown - Chemistry: The Central Science 14th Edition

Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232당신이 사용하는 게 아니라요?교과서 변경

Brown 14th Edition

Ch.6 - Electronic Structure of Atoms

문제 41b

6장, 문제 41b

The visible emission lines observed by Balmer all involved nf = 2. (b) Calculate the wavelengths of the first three lines in the Balmer series—those for which ni = 3, 4, and 5—and identify these lines in the emission spectrum shown in Figure 6.11.

검증된 단계별 안내

Identify the formula to calculate the wavelength of the emission lines in the Balmer series. The formula used is the Rydberg formula for hydrogen: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \), where \( \lambda \) is the wavelength, \( R_H \) is the Rydberg constant (approximately 1.097 x 10^7 m^-1), \( n_f \) is the final energy level (2 for Balmer series), and \( n_i \) is the initial energy level.

Plug in the values for the first line where \( n_i = 3 \) and \( n_f = 2 \) into the Rydberg formula to calculate the reciprocal of the wavelength (1/\( \lambda \)).

Repeat the calculation for the second line where \( n_i = 4 \).

Repeat the calculation for the third line where \( n_i = 5 \).

Convert the reciprocal wavelengths obtained in the previous steps to actual wavelengths by taking the reciprocal of each result. These wavelengths correspond to the first three lines in the Balmer series.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Balmer Series

The Balmer series refers to the set of spectral lines corresponding to electron transitions from higher energy levels (ni) to the second energy level (nf = 2) in a hydrogen atom. These transitions result in the emission of visible light, and the wavelengths of these lines can be calculated using the Rydberg formula. The first three lines in this series correspond to transitions from ni = 3, 4, and 5.

Rydberg Formula

The Rydberg formula is a mathematical equation used to predict the wavelengths of spectral lines in hydrogen and other hydrogen-like atoms. It is expressed as 1/λ = R_H (1/nf^2 - 1/ni^2), where R_H is the Rydberg constant, λ is the wavelength, nf is the final energy level, and ni is the initial energy level. This formula is essential for calculating the wavelengths of the lines in the Balmer series.

Emission Spectrum

An emission spectrum is a spectrum of the electromagnetic radiation emitted by a source, which in this case is the hydrogen atom. When electrons transition between energy levels, they emit photons at specific wavelengths, resulting in distinct lines on the spectrum. Identifying these lines allows for the analysis of the energy levels of the atom and the characteristics of the emitted light.

관련 실천

교과서 질문

One of the emission lines of the hydrogen atom has a wavelength of 94.974 nm. (a) In what region of the electromagnetic spectrum is this emission found?

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교과서 질문

(b) Is this line in the visible region of the electromagnetic spectrum?

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교과서 질문

The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed.

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교과서 질문

The visible emission lines observed by Balmer all involved nf = 2. (a) Which of the following is the best explanation of why the lines with nf = 3 are not observed in the visible portion of the spectrum: (i) Transitions to nf = 3 are not allowed to happen, (ii) transitions to nf = 3 emit photons in the infrared portion of the spectrum, (iii) transitions to nf = 3 emit photons in the ultraviolet portion of the spectrum, or (iv) transitions to nf = 3 emit photons that are at exactly the same wavelengths as those to nf = 2.

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교과서 질문

Consider a transition of the electron in the hydrogen atom from n = 8 to n = 3. (b) Will the light be absorbed or emitted?

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교과서 질문

The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. (b) Calculate the wavelengths of the first three lines in the Lyman series—those for which ni = 2, 3, and 4.

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