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Chapter 7: Problem 64

An excited hydrogen atom emits light with a wavelength of \(397.2 \mathrm{~nm}\) to reach the energy level for which \(n=2\). In which principal quantum level did the electron begin?

Short Answer

Expert verified
The electron initially started in the principal quantum level \(n_{2} = 3\).

Step by step solution

01

Recall the Rydberg formula

The Rydberg formula 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 emitted light, \(R_H\) is the Rydberg constant for hydrogen \((R_H = 1.097 \times 10^7 \, \text{m}^{-1})\), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.
02

Insert known values into the Rydberg formula

We know the final energy level \(n_1 = 2\), and we are given the wavelength of the emitted light \(\lambda = 397.2\, \text{nm}\). We need to find the initial energy level \(n_2\). Convert the given wavelength from nanometers to meters: \[\lambda = 397.2\, \text{nm} \times \frac{1\, \text{m}}{10^9\, \text{nm}} = 397.2 \times 10^{-9}\, \text{m}\] Now, we insert the known values into the Rydberg formula and solve for \(n_2\): \[\frac{1}{397.2\times10^{-9} \, \text{m}} = R_H \left(\frac{1}{4} - \frac{1}{n_{2}^2}\right)\]
03

Solve for the initial energy level, n2

Rearrange the equation and solve for \(n_{2}^2\): \[n_{2}^2 = \frac{1}{\frac{1}{R_H\lambda}+\frac{1}{4}}\]
04

Calculate n2, the initial energy level

Substitute the given values into the equation: \[n_{2}^2 = \frac{1}{\frac{1}{(1.097\times10^7 \, \text{m}^{-1})(397.2\times10^{-9} \, \text{m})}+\frac{1}{4}}\] Calculate \(n_{2}^2\): \[n_{2}^2 \approx 6.003\] Since the principal quantum level must be a positive integer, we round up to the nearest whole number: \[n_{2} = 3\] So, the electron initially started in the principal quantum level \(n_{2} = 3\).

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Most popular questions from this chapter

Answer the following questions assuming that \(m_{s}\) could have three values rather than two and that the rules for \(n, \ell\), and \(m_{\ell}\) are the normal ones. a. How many electrons would an orbital be able to hold? b. How many elements would the first and second periods in the periodic table contain? c. How many elements would be contained in the first transition metal series? d. How many electrons would the set of 4 forbitals be able to hold?

Are the following statements true for the hydrogen atom only, true for all atoms, or not true for any atoms? a. The principal quantum number completely determines the energy of a given electron. b. The angular momentum quantum number, \(\ell\), determines the shapes of the atomic orbitals. c. The magnetic quantum number, \(m_{\ell}\), determines the direction that the atomic orbitals point in space.

One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for \(\mathrm{Li}, \mathrm{N}, \mathrm{Ni}, \mathrm{Te}, \mathrm{Ba}\), and \(\mathrm{Hg} .\) Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=3 \rightarrow n=2\) b. \(n=4 \rightarrow n=2\) c. \(n=2 \rightarrow n=1\)

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. Write the formula for a possible oxide and a possible oxyanion of element 106 .
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