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Chapter 7: Problem 154

Calculate, to four significant figures, the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the \(n=5\) state and then fall to states with smaller values of \(n\).

Short Answer

Expert verified
The longest wavelength of light emitted by electrons in the hydrogen atom falling from the \(n=5\) state to smaller values of \(n\) is \(656.7 \, \mathrm{nm}\) and the shortest wavelength is \(121.5 \, \mathrm{nm}\), both calculated to four significant figures.

Step by step solution

01

Introduce the Balmer-Rydberg Formula

The Balmer-Rydberg formula is given by: \[ \frac{1}{\lambda} = R_\mathrm{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \(位\) is the wavelength of the emitted light, \(R_\mathrm{H}\) is the Rydberg constant for hydrogen (\(1.097 \times 10^7 \mathrm{m}^{-1}\)), \(n_1\) is the lower energy state (final state), and \(n_2\) is the higher energy state (initial state). Note that \(n_1 < n_2\).
02

Calculate the longest wavelength (n=4)

For the longest wavelength, the electron will fall from \(n_2 = 5\) to the nearest lower state, \(n_1 = 4\). Plug these values into the formula: \[ \frac{1}{\lambda_\mathrm{longest}} = (1.097 \times 10^7) \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \] and solve for \(位_\mathrm{longest}\): \[ \lambda_\mathrm{longest} = \frac{1}{(1.097 \times 10^7) \left( \frac{1}{16} - \frac{1}{25} \right)} = 6.567 \times 10^{-7} \mathrm{m} \] To four significant figures, the longest wavelength is \(656.7 \, \mathrm{nm}\).
03

Calculate the shortest wavelength (n=1)

For the shortest wavelength, the electron will fall from \(n_2 = 5\) all the way down to the lowest energy state, \(n_1 = 1\). Plug these values into the formula: \[ \frac{1}{\lambda_\mathrm{shortest}} = (1.097 \times 10^7) \left( \frac{1}{1^2} - \frac{1}{5^2} \right) \] and solve for \(位_\mathrm{shortest}\): \[ \lambda_\mathrm{shortest} = \frac{1}{(1.097 \times 10^7) \left( 1 - \frac{1}{25} \right)} = 1.215 \times 10^{-7} \mathrm{m} \] To four significant figures, the shortest wavelength is \(121.5 \, \mathrm{nm}\).
04

Conclusion

The longest wavelength of light emitted by electrons in the hydrogen atom falling from the \(n=5\) state to smaller values of \(n\) is \(656.7 \, \mathrm{nm}\) and the shortest wavelength is \(121.5 \, \mathrm{nm}\), both calculated to four significant figures.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Atom
The hydrogen atom is the simplest atom consisting of just one proton and one electron. This simplicity makes it a perfect model to understand various quantum mechanics concepts.
In a hydrogen atom, the electron revolves around the proton in discrete orbits or energy levels. Each of these energy levels is represented by a principal quantum number denoted by \(n\).
The higher the value of \(n\), the farther the orbit is from the nucleus, and the higher the potential energy of the electron. However, electrons can jump between these energy levels, either absorbing or emitting energy in the form of light.
Electron Transition
When an electron in a hydrogen atom transitions from a higher energy level (\(n_2\)) to a lower energy level (\(n_1\)), it emits energy. This energy corresponds to a specific wavelength of light.
This process is known as electron transition. Understanding electron transitions is essential, as they are the basis for phenomena like spectral lines.
Excited electrons release energy when they move to a lower state. The difference in energy between the two levels determines the light's wavelength.
  • Higher energy difference results in light of shorter wavelength (higher frequency).
  • Lower energy difference results in light of longer wavelength (lower frequency).
These transitions are quantized, meaning electrons can only move between specific energy levels.
Wavelength Calculation
Wavelength calculation in electronic transitions involves determining the specific wavelength of light emitted during an electron's transition between two energy levels.
The formula used for this calculation is known as the Balmer-Rydberg formula.
By using this formula, one can calculate the exact value of the wavelength by knowing the initial and final quantum numbers, \(n_2\) and \(n_1\), respectively.
  • Start with the Balmer-Rydberg formula:
    \[ \frac{1}{\lambda} = R_\mathrm{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
  • Substitute the Rydberg constant for hydrogen (
    \(R_\mathrm{H} = 1.097 \times 10^7\, \mathrm{m}^{-1}\)
  • Solve for \(\lambda\), the wavelength of emitted light.
This calculation provides insight into the energy levels involved and the nature of light emissions in hydrogen atoms.
Rydberg Constant
The Rydberg constant, denoted by \(R_\mathrm{H}\), is a fundamental constant used to describe the wavelengths of spectral lines of hydrogen.
It is significant in the Balmer-Rydberg formula used to calculate the wavelengths of light emitted by electron transitions in the hydrogen atom.
For hydrogen, the value of the Rydberg constant is approximately \(1.097 \times 10^7 \, \mathrm{m}^{-1}\).
  • This constant links the energy levels and wavelengths of emitted light.
  • It forms the basis of calculating spectral lines for hydrogen, helping to understand quantum levels.
The Rydberg constant is central to understanding energy transitions in hydrogen and indicates how quantum mechanics governs atomic structures.

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Most popular questions from this chapter

Which of the following statements is(are) true? a. \(\mathrm{F}\) has a larger first ionization energy than does \(\mathrm{Li}\). b. Cations are larger than their parent atoms. c. The removal of the first electron from a lithium atom (electron configuration is \(1 s^{2} 2 s^{1}\) ) is exothermic - that is, removing this electron gives off energy. d. The He atom is larger than the \(\mathrm{H}^{+}\) ion. e. The \(\mathrm{Al}\) atom is smaller than the \(\mathrm{Li}\) atom.

How many unpaired electrons are present in each of the following in the ground state: \(\mathrm{O}, \mathrm{O}^{+}, \mathrm{O}^{-}, \mathrm{Os}, \mathrm{Zr}, \mathrm{S}, \mathrm{F}\), Ar?

Element 106 has been named seaborgium, \(\mathrm{Sg}\), in honor of Glenn Seaborg, discoverer of the first transuranium element. a. Write the expected electron configuration for element 106 . b. What other element would be most like element 106 in its properties? c. Predict the formula for a possible oxide and a possible oxyanion of element 106 .

Consider the following approximate visible light spectrum: Barium emits light in the visible region of the spectrum. If each photon of light emitted from barium has an energy of \(3.59 \times 10^{-19} \mathrm{~J}\), what color of visible light is emitted?

The following numbers are the ratios of second ionization energy to first ionization energy: Na: \(\quad 9.2\) P: \(1.8\) Mg: \(\quad 2.0\) \(\mathrm{S}: \quad 2.3\) Al: \(\quad 3.1\) \(\mathrm{Cl}: \quad 1.8\) Si: \(\quad 2.0\) Ar: \(1.8\) Explain these relative numbers.
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