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Chapter 6: Problem 100

The Chemistry and Life box in Section 6.7 described the techniques called NMR and MRI. (a) Instruments for obtaining MRI data are typically labeled with a frequency, such as 600 MHz. In what region of the electromagnetic spectrum does a photon with this frequency belong? (b) What is the value of \(\Delta E\) in Figure 6.27 that would correspond to the absorption of a photon of radiation with frequency \(450 \mathrm{MHz} ?(\mathbf{c})\) When the 450 -MHz photon is absorbed, does it change the spin of the electron or the proton on a hydrogen atom?

Short Answer

Expert verified
(a) A photon with a frequency of 600 MHz belongs to the radio wave region of the electromagnetic spectrum, specifically in the UHF (Ultra High Frequency) band. (b) The energy change (鈭咵) for the absorption of a photon with a frequency of 450 MHz is 2.982 脳 10鈦宦测伒 J. (c) The absorption of a 450 MHz photon changes the spin of the proton, not the electron, on a hydrogen atom.

Step by step solution

01

(a) Determine photon energy and spectrum region

First, we need to determine the energy of a photon with a frequency of 600 MHz. Use the formula: \[E = h\nu\] Where \(E\) is the photon energy, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{J\:s}\)), and \(\nu\) is the frequency. Remember to change MHz to Hz by multiplying by \(10^{6}\). \[E = (6.626 \times 10^{-34} \mathrm{J\:s}) (600 \times 10^{6} \mathrm{Hz})\] \[E = 3.976 \times 10^{-25} \mathrm{J}\] Now, we need to determine the region in the electromagnetic spectrum where this energy belongs. The 600 MHz frequency corresponds to radio waves, specifically, it falls in the UHF (Ultra High Frequency) band.
02

(b) Calculate energy change for the 450 MHz photon

Now we need to calculate the energy change (\(\Delta E\)) for the absorption of a photon with a frequency of 450 MHz. Use the same formula: \[\Delta E = h\nu\] \[\Delta E = (6.626 \times 10^{-34} \mathrm{J\:s}) (450 \times 10^{6} \mathrm{Hz})\] \[\Delta E = 2.982 \times 10^{-25} \mathrm{J}\]
03

(c) Determine the effect of the 450 MHz photon on electron or proton spin

Nuclear Magnetic Resonance (NMR) and Magnetic Resonance Imaging (MRI) techniques are based on the interaction of photons with nuclei (protons), not electrons. When a photon is absorbed, it affects the spin states of nuclei (protons) in the magnetic field. Thus, the absorption of a 450 MHz photon changes the spin of the proton, not the electron, on a hydrogen atom.

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Most popular questions from this chapter

Consider a transition in which the hydrogen atom is excited from \(n=1\) to \(n=\infty\). (a) What is the end result of this transition? (b) What is the wavelength of light that must be absorbed to accomplish this process? (c) What will occur if light with a shorter wavelength than that in part (b) is used to excite the hydrogen atom? (d) How the results of parts (b) and (c) related to the plot shown in Exercise \(6.88\) ?

Using the periodic table as a guide, write the condensed electron configuration and determine the number of unpaired electrons for the ground state of (a) \(\mathrm{Br}\), (b) Ga, (c) Hf, (d) Sb, (e) Bi, (f) Sg.

The visible emission lines observed by Balmer all involved \(n_{\mathrm{f}}=2\). (a) Which of the following is the best explanation of why the lines with \(n_{\mathrm{f}}=3\) are not observed in the visible portion of the spectrum: (i) Transitions to \(n_{\mathrm{f}}=3\) are not allowed to happen, (ii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the infrared portion of the spectrum, (iii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the ultraviolet portion of the spectrum, or (iv) transitions to \(n_{\mathrm{f}}=3\) emit photons that are at exactly the same wavelengths as those to \(n_{\mathrm{f}}=2\). (b) Calculate the wavelengths of the first three lines in the Balmer series-those for which \(n_{\mathrm{i}}=3,4\), and 5 -and identify these lines in the emission spectrum shown in Figure 6.11.

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