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Hydrogen atoms are placed in an external magnetic field. The protons can make transitions between states in which the nuclear spin component is parallel and antiparallel to the field by absorbing or emitting a photon. What magnetic- field magnitude is required for this transition to be induced by photons with frequency 22.7 \(\mathrm{MHz} ?\)

Short Answer

Expert verified
Required magnetic field is approximately 0.53 Tesla.

Step by step solution

01

Understand the Problem Context

We need to determine the magnetic field strength required to induce transitions in hydrogen nuclei when exposed to photons of a specific frequency (22.7 MHz). The transitions occur between states where the nuclear spin is parallel and antiparallel to the field.
02

Identify Relevant Physics

The energy difference between the two spin states of the proton in a magnetic field can be expressed using the equation: \(\Delta E = h u\), where \( u \) is the frequency of the associated photon. Also, the energy difference is related to the magnetic field \(B\) by \(\Delta E = 2\mu_B B\), where \(\mu_B\) is the magnetic moment of the proton.
03

Relate Energy Difference and Magnetic Field

Since \(\Delta E = h u = 2 \mu_B B\), we can express the magnetic field \(B\) in terms of \(u\): \[ B = \frac{h u}{2 \mu_B} \]We know \(h\) (Planck's constant) is approximately \(6.626 \times 10^{-34} \text{ J s}\), and \(\mu_B\) is approximately \(1.4105 \times 10^{-26} \text{ J T}^{-1}\).
04

Calculate the Magnetic Field Strength

Insert the given photon frequency (\(22.7 \text{ MHz} = 22.7 \times 10^6 \text{ Hz}\)) into the equation:\[ B = \frac{6.626 \times 10^{-34} \times 22.7 \times 10^6}{2 \times 1.4105 \times 10^{-26}} \]Perform the calculation to find:\[ B \approx 0.53 \text{ T} \]
05

Conclusion

The magnetic field required to induce the proton transition with a 22.7 MHz frequency photon is approximately 0.53 Tesla.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Atom Transitions
In the study of atomic physics, understanding how hydrogen atoms transition between different energy states is fundamental. These transitions occur when photons - light particles - interact with the atoms.

When a hydrogen atom is in a magnetic field, its nucleus, particularly the protons, can exist in different spin states. The two main states include:
  • Parallel spin state: where the nuclear spin aligns with the direction of the magnetic field.
  • Antiparallel spin state: where the nuclear spin opposes the direction of the field.
The process of flipping between these states involves transitions that require or release energy. This energy corresponds to the absorption or emission of photons, often of specific frequencies.

Such transitions are not just theoretical; they have practical applications. For instance, magnetic resonance imaging (MRI) exploits these principles to visualize internal structures of the human body.
Nuclear Spin States
Nuclear spin states refer to how the nucleus's intrinsic angular momentum is oriented, specifically in the presence of an external magnetic field. Protons, being particles with a spin, will naturally orient themselves in relation to the magnetic lines of force.

In the context of a magnetic field, a hydrogen nucleus can exhibit:
  • Spin parallel to the magnetic field, which is a lower energy state.
  • Spin antiparallel to the magnetic field, which is a higher energy state.
These spin states are quantized, meaning they only take on specific values. Transitions between these states necessitate the absorption of energy (via photons) when moving to a higher energy state or emission of energy when falling to a lower energy state. Understanding these spin states is crucial, as it forms the basis for technologies like Nuclear Magnetic Resonance (NMR) used in chemistry and medicine.
Photon Frequency
Photon frequency is a crucial concept when discussing transitions in atomic systems like hydrogen. A photon has both energy and a corresponding frequency, and these properties are directly linked through Planck's law, expressed as \[E = h \cdot u\]where \(E\) is energy, \(h\) is Planck's constant, and \(u\) (or \(u\)) is the frequency.

The frequency of a photon determines whether its energy is suitable for interacting with particles like protons, inducing transitions. In the case of our hydrogen atom example, a photon with a frequency of 22.7 MHz is necessary to cause a transition between spin states.

Higher frequency photons carry more energy, suitable for inducing changes in atomic structures, whereas lower frequency photons have less energy. This relationship underscores many fields, such as spectroscopy, where the frequency of absorbed or emitted light gives clues about atomic structures and interactions.
Energy Difference
The energy difference between two nuclear spin states is at the core of understanding atomic transitions in a magnetic field. In a magnetic field, the energy difference \(\Delta E\) between the parallel and antiparallel spin states is related to the magnetic field strength and the magnetic moment of the proton.

This relationship can be captured through this equation:\[\Delta E = 2 \mu_B B\]Here, \(\Delta E\) represents the energy needed to flip the spin state, \(B\) is the magnetic field, and \(\mu_B\) is the magnetic moment.

The necessity for a photon with a specific frequency arises from this energy difference. The photon's energy must match \(\Delta E\) to facilitate a transition.

This linkage forms the basis for determining the magnetic field required for specific transitions, exemplified by our exercise which calculates a magnetic field strength allowing for transitions with photons of 22.7 MHz frequency. Understanding these energy differences is vital in manipulating and measuring atomic states through technologies like spectroscopy and magnetic imaging.

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Most popular questions from this chapter

Measurements indicate that 27.83\(\%\) of all rubidium atoms currently on the earth are the radioactive \(^{87} \mathrm{Rb}\) isotope. The rest are the stable \(^{87} \mathrm{Rb}\) isotope. The half-life of \(^{87} \mathrm{Rb}\) is \(4.75 \times 10^{10} \mathrm{y}\) . Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^{87} \mathrm{Rb}\) when our solar system was formed \(4.6 \times 10^{9} \mathrm{y}\) ago?

What particle \((\alpha\) particle, electron, or positron) is emitted in the following radioactive decays? (a) \(_{14}^{27} \mathrm{Si} \rightarrow_{13}^{27} \mathrm{Al}\) (b) \(^{238} \mathrm{U} \rightarrow_{90}^{234} \mathrm{Th} ;\) (c) \(_{33}^{74} \mathrm{As} \rightarrow_{34}^{74} \mathrm{Se}\)

How many protons and how many neutrons are there in a nucleus of the most common isotope of (a) silicon, \(_{14}^{28} \mathrm{Si} ;\) (b) rubidium, \(\frac{85}{37} \mathrm{Rb} ;\) (c) thallium, \(_{81}^{205} \mathrm{Tl} ?\)

We Are Stardust. In 1952 spectral lines of the element technetium- 99\(\left(^{99} \mathrm{Tc}\right)\) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has \(no\) stable isotopes, and the half-life of \(^{99} \mathrm{Tc}\) is \(200,000\) years. (a) For how many half-lives has the \(^{99} \mathrm{Tc}\) been in the red-giant star if its age is 10 billion years? (b) What fraction of the original \(^{99} \mathrm{Tc}\) would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and other nuclear processes. If the \(^{99} \mathrm{Tc}\) had been part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This knowledge is what led the late astronomer Carl Sagan to proclaim that "we are stardust."

The common isotope of uranium, \(^{238} \mathrm{U},\) has a half-life of \(4.47 \times 10^{9}\) years, decaying to \(^{234} \mathrm{Th}\) by alpha emission. (a) What is the decay constant? (b) What mass of uranium is required for an activity of 1.00 curie? (c) How many alpha particles are emitted per second by 10.0 g of uranium?

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