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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 \(λ\) 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 \(λ_\mathrm{longest}\): \[ \lambda_\mathrm{longest} = \frac{1}{(1.097 \times 10^7) \left( \frac{1}{16} - \frac{1}{25} \right)} = 6.567 \times 10^{-7} \mathrm{m} \] 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 \(λ_\mathrm{shortest}\): \[ \lambda_\mathrm{shortest} = \frac{1}{(1.097 \times 10^7) \left( 1 - \frac{1}{25} \right)} = 1.215 \times 10^{-7} \mathrm{m} \] 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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Most popular questions from this chapter

Although no currently known elements contain electrons in \(g\) orbitals in the ground state, it is possible that these elements will be found or that electrons in excited states of known elements could be in \(g\) orbitals. For \(g\) orbitals, the value of \(\ell\) is \(4 .\) What is the lowest value of \(n\) for which \(g\) orbitals could exist? What are the possible values of \(m_{\ell} ?\) How many electrons could a set of \(g\) orbitals hold?

The first ionization energies of As and Se are \(0.947\) and \(0.941\) MJ/mol, respectively. Rationalize these values in terms of electron configurations.

Assume that we are in another universe with different physical laws. Electrons in this universe are described by four quantum numbers with meanings similar to those we use. We will call these quantum numbers \(p, q, r\), and \(s\). The rules for these quantum numbers are as follows: \(p=1,2,3,4,5, \ldots\) \(q\) takes on positive odd integers and \(q \leq p\). \(r\) takes on all even integer values from \(-q\) to \(+q\). (Zero is considered an even number.) \(s=+\frac{1}{2}\) or \(-\frac{1}{2}\) a. Sketch what the first four periods of the periodic table will look like in this universe. b. What are the atomic numbers of the first four elements you would expect to be least reactive? c. Give an example, using elements in the first four rows, of ionic compounds with the formulas XY, \(\mathrm{XY}_{2}, \mathrm{X}_{2} \mathrm{Y}, \mathrm{XY}_{3}\), and \(\mathrm{X}_{2} \mathrm{Y}_{3}\). d. How many electrons can have \(p=4, q=3 ?\) e. How many electrons can have \(p=3, q=0, r=0\) ? f. How many electrons can have \(p=6\) ?

Which of elements \(1-36\) have one unpaired electron in the ground state?

A particle has a velocity that is \(90 . \%\) of the speed of light. If the wavelength of the particle is \(1.5 \times 10^{-15} \mathrm{~m}\), calculate the mass of the particle.
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