Question

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?

Short 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.

Step-by-step solution

Determine the EM Spectrum region for the photon

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: \(λ = \frac{3.00 \times 10^8 \text{ m/s}}{6.00 \times 10^8 \text{ Hz}} = 0.50 \text{ m}\)

Identify the EM Spectrum region

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.

Calculate the energy difference for the photon

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.

Determine the changing spin after photon absorption

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.

Key concepts

Nuclear Magnetic Resonance (NMR)

Nuclear Magnetic Resonance (NMR) is a fascinating technique used to explore the properties of atomic nuclei. At its core, NMR relies on the fact that certain nuclei have a property called spin, which creates a magnetic moment. When these nuclei are placed in a magnetic field, they behave like tiny magnets.

NMR is widely used in chemistry and medicine. Here’s how it works:

• When placed in a magnetic field, nuclei have specific energy levels based on their spin orientation.
• By applying a frequency corresponding to the energy differences between these levels, nuclei can absorb that energy and "flip" their spin.
• This absorption can be measured and analyzed to give detailed information about the molecule’s structure.

This method is incredibly valuable as it allows scientists to determine the chemical environment of atoms within a molecule, leading to applications like MRI in medical imaging. It's amazing how NMR enables us to peek into the molecular world!

Radio Wave Region

The radio wave region of the electromagnetic spectrum is an intriguing place! This region encompasses electromagnetic waves with long wavelengths, typically from 1 millimeter to more than 100 kilometers, and low frequencies, ranging from 300 gigahertz (GHz) to 3 kilohertz (kHz).

Here’s why radio waves are special:

• They are used in numerous applications, from broadcasting radio and television signals to GPS and mobile phone communications.
• They have enough power to travel long distances, which is great for communication systems.
• In physics and chemistry, radio waves are used in NMR as they are capable of causing transitions in nuclear spin states.

With a frequency of 600 MHz, as seen in the NMR example, the electromagnetic waves fall squarely into the radio wave region. This classification is crucial because the less energetic radio waves are ideal for probing the nuclear characteristics without damaging the material under study.

Photon Energy

Photon energy is a fundamental concept in understanding how light interacts with matter. Each photon, which is a quantum of electromagnetic radiation, carries energy based on its frequency, given by the equation:\[ E = hu \]where \(E\) is the energy, \(h\) is Planck’s constant (\(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\)), and \(u\) is the frequency of the photon.

Why is this important?

• Photon energy determines how light can influence or interact with materials, playing a critical role in phenomena like photosynthesis or photoelectric effects.
• In NMR, low photon energies from the radio wave region cause nuclear spin transitions, altering the energy states of nuclei without the risk of ionizing the material.

With photons in the radio wave region, energy levels are low (like the 450-MHz example from the exercise). This specificity in energy is vital for applications where subtle changes in material characteristics need to be detected without causing structural damage. The interplay of frequency and energy enables technologies that uncover deep scientific insights.