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Chapter 36: Problem 33

The \(4 p \rightarrow 3 s\) transition in sodium produces a double spectral line at 330.2 and \(330.3 \mathrm{nm} .\) What's the energy splitting of the \(4 p\) level?

Short Answer

Expert verified
The energy splitting of the \(4 p\) level is calculated as the absolute difference between the energies associated with the given spectral lines, which is then converted into electron-volts for convenience.

Step by step solution

01

Calculate the Energy associated with each Wavelength

First, calculate the energy associated with each line using Planck's relation: \(E = h \cdot f\), where \(h\) is Planck's constant and \(f\) is the frequency. Frequency and wavelength \(\lambda\) are related by the speed of light \(c\) as \(f = c/\lambda\) . Therefore, we can express the energy \(E\) as a function of wavelength: \(E = h \cdot c/\lambda\) . Use this equation to calculate the energies for the given spectral lines (330.2 nm and 330.3 nm). The values of Planck's constant \(h\) and the speed of light \(c\) are standard physical constants.
02

Calculate the Energy Difference

The energy difference or 'splitting' between the two levels is given by the absolute difference of the two energy values calculated in Step 1. Subtract the energy of the spectral line at 330.3 nm from the energy of the spectral line at 330.2 nm to obtain the energy splitting.
03

Convert to Appropriate Units

The result obtained in Step 2 will be in Joules, which might be a very small number. It might be useful to convert this result into more commonly used units in atomic physics, such as electron-volts (eV). 1eV = \(1.602 \times 10^{-19}\) Joules.

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Most popular questions from this chapter

Show that the wavelength \(\lambda\) in \(\mathrm{nm}\) of a photon with energy \(E\) in eV is \(\lambda=1240 / E\)

Adapt part (b) of Example 36.1 to find the probability that an electron in the hydrogen ground state will be found beyond two Bohr radii.

A hydrogen atom is in the \(2 s\) state. Find the probability that its electron will be found (a) beyond one Bohr radius and (b) beyond 10 Bohr radii.

Use shell notation to characterize rubidium's outermost electron.

With sufficient energy, it's possible to eject an electron from an inner atomic orbital. A higher-energy electron will then drop into the unoccupied state, emitting a photon with energy equal to the difference between the two levels. For inner-shell electrons, photon energies are in the keV range, putting them in the X-ray region of the spectrum. These characteristic X rays are labeled with the letter indicating the shell to which the electron drops, followed by a Greek letter indicating the higher level from which it drops; thus \(K \alpha\) designates a transition from the \(L\) shell to the \(K\) shell. Characteristic X rays provide scientists and physicians with an important diagnostic tool. Environmental scientists bombard pollution samples with high- energy electrons, knocking out inner-shell electrons and thus producing X-ray spectra that help identify contaminants (Fig. \(36.20 a\) ). Geologists do the same with rocks. Medical radiologists reverse the process, exploiting the fact that X rays cause inner-shell transitions as well as complete ejection of inner-shell electrons. In particular, radiologists use the element barium in this way to produce high-contrast X-ray images of the intestinal tract \((\text { Fig. } 36.20 b)\)(GRAPH CANNOT COPY) (a) An \(\mathrm{X}\) -ray spectrum from air pollutants trapped on a filter. The labeled peaks show the presence of lead (Pb) and arsenic (As), as evidenced by \(K \alpha, K \beta, L \alpha,\) and \(L \beta\) characteristic X rays. (b) \(\mathrm{X}\) -ray of an intestinal tract, made by coating the intestinal wall with X-ray-opaque barium Molybdenum's X-ray spectrum has its \(K \alpha\) peak at 17.4 keV. The corresponding X-ray wavelength is closest to a. \(1 \mathrm{pm}\) b. \(100 \mathrm{pm}\) c. \(1 \mathrm{nm}\) d. \(100 \mathrm{nm}\)
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