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Chapter 2: Problem 151

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 when transitioning from the \(n=5\) state to a smaller value of \(n\) is approximately \(6.564 \times 10^{-7}\ m\), and the shortest wavelength is approximately \(4.342 \times 10^{-7}\ m\).

Step by step solution

01

Recall the Rydberg Formula

The Rydberg formula is used to calculate the wavelengths of light emitted during electron transitions in hydrogen. The formula is given by: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_j^2} - \frac{1}{n_i^2}\right)\] Here, \(\lambda\) is the wavelength of light emitted, \(R_H\) is the Rydberg constant for hydrogen (\(R_H \approx 1.097\times10^7 m^{-1}\)), \(n_i\) is the initial quantum state, and \(n_j\) is the final quantum state.
02

Calculate the longest wavelength

To calculate the longest wavelength, we'll use the values \(n_i = 5\) and \(n_j = 4\) in the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{4^2} - \frac{1}{5^2}\right)\] Now, solve for \(\lambda\): \[\lambda = \frac{1}{R_H \left(\frac{1}{4^2} - \frac{1}{5^2}\right)}\] Plug in the value of \(R_H\) and solve for \(\lambda\): \[\lambda \approx \frac{1}{1.097\times10^7 \left(\frac{1}{16} - \frac{1}{25}\right)} \approx 6.564\times10^{-7} m\] The longest wavelength the electron can emit, when falling from \(n=5\) to \(n=4\), is approximately \(6.564\times10^{-7} m\).
03

Calculate the shortest wavelength

To calculate the shortest wavelength, we'll use the values \(n_i = 5\) and \(n_j = 1\) in the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{1^2} - \frac{1}{5^2}\right)\] Now, solve for \(\lambda\): \[\lambda = \frac{1}{R_H \left(\frac{1}{1^2} - \frac{1}{5^2}\right)}\] Plug in the value of \(R_H\) and solve for \(\lambda\): \[\lambda \approx \frac{1}{1.097\times10^7 \left(\frac{1}{1} - \frac{1}{25}\right)} \approx 4.342\times10^{-7} m\] The shortest wavelength the electron can emit, when falling from \(n=5\) to \(n=1\), is approximately \(4.342\times10^{-7} m\).
04

Express the results to four significant figures

Express the results of the longest and shortest wavelengths to four significant figures: \[\lambda_{longest} \approx 6.564\times10^{-7} m \approx 6.564 \times 10^{-7}\ m\] \[\lambda_{shortest} \approx 4.342\times10^{-7} m \approx 4.342 \times 10^{-7}\ m\] Thus, the longest wavelength of light emitted is approximately \(6.564 \times 10^{-7}\ m\) and the shortest wavelength is approximately \(4.342 \times 10^{-7}\ m\).

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Most popular questions from this chapter

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

Valence electrons are those electrons in the outermost principal quantum level (highest \(n\) level) of an atom in its ground state. Groups \(1 \mathrm{A}\) to \(8 \mathrm{A}\) have from 1 to 8 valence electrons. For each group of the representative elements (1A-8A), give the number of valence electrons, the general valence electron configuration, a sample element in that group, and the specific valence electron configuration for that element.

Assume that a hydrogen atom's electron has been excited to the \(n=6\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

The successive ionization energies for an unknown element are \(I_{1}=896 \mathrm{kJ} / \mathrm{mol}\) \(\overline{I_{2}}=1752 \mathrm{kJ} / \mathrm{mol}\) \(I_{3}=14,807 \mathrm{kJ} / \mathrm{mol}\) \(I_{4}=17,948 \mathrm{kJ} / \mathrm{mol}\) To which family in the periodic table does the unknown element most likely belong?

The wave function for the \(2 p_{z}\) orbital in the hydrogen atom is $$ \psi_{2 p_{i}}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\alpha / 2} \cos \theta $$ where \(a_{0}\) is the value for the radius of the first Bohr orbit in meters \(\left(5.29 \times 10^{-11}\right), \sigma\) is \(Z\left(r / a_{0}\right), r\) is the value for the distance from the nucleus in meters, and \(\theta\) is an angle. Calculate the value of \(\psi_{2 p^{2}}\) at \(r=a_{0}\) for \(\theta=0^{\circ}\left(z \text { axis) and for } \theta=90^{\circ}\right.\) (xy plane).
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