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Ch.6 - Electronic Structure of Atoms
Chapter 6, Problem 39b

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

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1
Understand the electromagnetic spectrum: The electromagnetic spectrum includes all types of electromagnetic radiation, ranging from gamma rays to radio waves. The visible region is a small part of this spectrum, typically ranging from wavelengths of about 380 nm to 750 nm.
Identify the wavelength of the line in question: To determine if a line is in the visible region, you need to know its wavelength. This information is usually provided in the problem or can be calculated using other given data.
Compare the wavelength to the visible range: Once you have the wavelength, compare it to the visible range (380 nm to 750 nm). If the wavelength falls within this range, the line is in the visible region.
Consider the context of the problem: Sometimes, the problem might involve transitions in atomic spectra, such as those of hydrogen. In such cases, specific series like the Balmer series are known to produce lines in the visible region.
Conclude based on the comparison: If the wavelength is within the visible range, you can conclude that the line is in the visible region of the electromagnetic spectrum. Otherwise, it is outside the visible region.

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Is energy emitted or absorbed when the following electronic transitions occur in hydrogen? (b) from an orbit of radius 0.846 nm to one of radius 0.212 nm

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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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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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