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Chapter 2: Problem 67

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Short Answer

Expert verified
In the given problem, an excited hydrogen atom emits light with a frequency of \(6.90 \times 10^{14} \, s^{-1}\), and we need to find the principal quantum level for the final state. By converting the frequency to wavelength (\(\lambda \approx 4.35 \times 10^{-7}\, m\)) and using the Rydberg formula, we can determine the final principal quantum number. The calculations result in the final principal quantum number \(n_2\) to be approximately \(5.15\). As principal quantum numbers must be integers, we round it to the nearest integer, so the final principal quantum level is \(n_2 = 5\).

Step by step solution

01

Recall the Rydberg formula for hydrogen

The Rydberg formula for hydrogen is given by: \[ \frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] where \( \lambda \) is the wavelength of the emitted light, \(R_H\) is the Rydberg constant for hydrogen (approximately \(1.097 \times 10^7 \, m^{-1}\)), \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final states respectively, with \(n_1 < n_2\).
02

Convert the frequency to wavelength

We are given the frequency of the emitted light, but we need the wavelength to use the Rydberg formula. We can convert the frequency to wavelength using the following formula: \[ v = \frac{c}{\lambda} \] where \(v\) is the frequency of the light, and \(c\) is the speed of light (approximately \(3.0\times 10^8 \, m/s\)). Now we can calculate the wavelength of the light: \[ \lambda = \frac{c}{v} = \frac{3.0 \times 10^8 \, m/s}{6.9 \times 10^{14} \, s^{-1}} \approx 4.35 \times 10^{-7}\, m \]
03

Apply the Rydberg formula

Now we can substitute the calculated wavelength and principal quantum number of the initial state (n=5) into the Rydberg formula and solve for the final principal quantum number (\(n_2\)): \[ \frac{1}{4.35 \times 10^{-7}\, m} = 1.097 \times 10^7 \, m^{-1} \left(\frac{1}{5^2} - \frac{1}{n_2^2}\right) \]
04

Solve for the final principal quantum number

We will now isolate the term \(\frac{1}{n_2^2}\) and then solve for \(n_2\): \[ \frac{1}{n_2^2} = \frac{1}{5^2} - \frac{1}{ 4.35 \times 10^{-7}\, m \times 1.097 \times 10^7 \, m^{-1}} \approx 0.038 \] Taking the reciprocal and then the square root of both sides, we get: \[ n_2 \approx \frac{1}{\sqrt{0.038}} \approx 5.15 \] Finally, as the principal quantum number must be an integer, we round it to the nearest integer: \[ n_2 = 5 \] So the final principal quantum level for this electronic transition is 5.

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Most popular questions from this chapter

In defining the sizes of orbitals, why must we use an arbitrary value, such as \(90 \%\) of the probability of finding an electron in that region?

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=4\) b. \(n=5, m_{\ell}=+1\) c. \(n=5, m_{s}=+\frac{1}{2}\) d. \(n=3, \ell=2\) e. \(n=2, \ell=1\)

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

Determine the maximum number of electrons that can have each of the following designations: \(2 f, 2 d_{x y}, 3 p, 5 d_{y z},\) and \(4 p\).

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.
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