The Chemistry and Life boxin 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 \(\mathrm{MHz}\) ? (c) When the 450 -MHz photon is absorbed, does it change the spin of the electron or the proton on a hydrogen atom?

A photon with a frequency of 600 MHz is given. To determine which region of the electromagnetic spectrum it belongs to, let's first convert the frequency to wavelength using the formula: \(λ = \frac{c}{ν}\) where \(λ\) is the wavelength, \(c\) is the speed of light (\(3.00 \times 10^8 \text{ m/s}\)), and \(ν\) is the frequency (600 MHz). We need to convert the frequency from MHz to Hz: \( ν = 600 \text{ MHz} \times 10^6 = 6.00 \times 10^8 \text{ Hz}\) Now, we can find the wavelength: \(λ = \frac{3.00 \times 10^8 \text{ m/s}}{6.00 \times 10^8 \text{ Hz}} = 0.50 \text{ m}\)

With the calculated wavelength of 0.50 meters, we can identify the region of the electromagnetic spectrum to which the photon belongs: This wavelength falls into the radio wave region of the electromagnetic spectrum.

We are given a photon frequency of 450 MHz and need to find the corresponding energy difference (\(\Delta E\)). To determine the energy of the photon, we can use the formula: \(E = hν\) where \(E\) is the energy, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\)), and \(ν\) is the frequency. First, let's convert the frequency to Hz: \( ν = 450 \text{ MHz} \times 10^6 = 4.50 \times 10^8\text{ Hz}\) Now, we can calculate the energy of the photon: \(E = (6.626 \times 10^{-34} \text{ J} \cdot \text{s})(4.50 \times 10^8 \text{ Hz}) = 2.98 \times 10^{-25} \text{ J}\) This energy difference corresponds to the absorption of the photon.

We need to find out if the electron or proton spin on a hydrogen atom changes when a 450-MHz photon is absorbed. Since the frequency given in the problem belongs to the radio wave region of the electromagnetic spectrum, and the energy difference is relatively low, the energy required to flip the spin is mainly associated with the proton (nuclear) spin, which is less sensitive to external electromagnetic fields. Therefore, the proton (nuclear) spin on the hydrogen atom changes when the 450-MHz photon is absorbed. #Answer#: (a) The photon with a frequency of 600 MHz belongs to the radio wave region of the electromagnetic spectrum. (b) The value of \(\Delta E\) corresponding to the absorption of a photon with a frequency of 450 MHz is \(2.98 \times 10^{-25}\) J. (c) When the 450-MHz photon is absorbed, the spin of the proton on a hydrogen atom changes.