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Chapter 43: Problem 3

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 MHz?

Short Answer

Expert verified
The required magnetic field magnitude is approximately 0.53 T.

Step by step solution

01

Understanding the Relationship

The transition between the spin states of protons in a magnetic field involves energy changes related to the frequency of the photon. The formula that links the energy difference with frequency is given by Planck's equation: \[ \Delta E = h \cdot u \]where \( \Delta E \) is the energy difference, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J} \cdot \text{s}) \), and \( u \) is the frequency.
02

Zeeman Effect Clarification

The energy difference \( \Delta E \) between the parallel and antiparallel states in a magnetic field \( B \) is given by:\[ \Delta E = 2 \cdot \mu_p \cdot B \]where \( \mu_p \) is the magnetic moment of the proton \( (1.41 \times 10^{-26} \text{ J/T}) \).
03

Equating the Energy Changes

By equating the expressions from Step 1 and Step 2, we get:\[ h \cdot u = 2 \cdot \mu_p \cdot B \]We can rearrange this formula to solve for the magnetic field \( B \): \[ B = \frac{h \cdot u}{2 \cdot \mu_p} \]
04

Substituting Known Values

Substitute the known values into the equation:Given \( u = 22.7 \times 10^6 \text{ Hz} \) and Planck's constant \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \), and \( \mu_p = 1.41 \times 10^{-26} \text{ J/T} \), we find:\[ B = \frac{6.626 \times 10^{-34} \times 22.7 \times 10^6}{2 \times 1.41 \times 10^{-26}} \]
05

Performing the Calculation

Calculate the magnetic field:\[ B = \frac{6.626 \times 22.7 \times 10^{-28}}{2.82 \times 10^{-26}} \]Solving this gives:\[ B \approx 0.53 \text{ T} \]

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Key Concepts

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

Magnetic Field
A magnetic field is a fundamental concept in physics representing a region where a magnetic force acts on charged particles. In this problem, hydrogen atoms placed in an external magnetic field experience effects on their protons. Protons, being positively charged particles, respond to the magnetic field by altering their spin states. Generally, a magnetic field can influence materials and particles in various ways:
  • It aligns magnetic dipoles.
  • Can induce electric currents in conductors.
  • Is quantified by its strength and direction.
Here, the magnetic field induces transitions in the proton's spin states, which we calculate by understanding both energy differences and photon interactions. The stronger the field, the greater the energy differences in spin states.
Proton Spin States
Protons possess an intrinsic property called 'spin,' which is similar to the idea of angular momentum. In the context of this exercise, protons in hydrogen atoms can have their spin aligned either parallel or antiparallel to the external magnetic field:
  • Parallel spin state: When the spin of the proton aligns with the magnetic field direction.
  • Antiparallel spin state: When the proton's spin is opposite to the direction of the magnetic field.
These different spin orientations correspond to slightly different energy levels. The Zeeman Effect is a crucial phenomenon observing how these energy levels split under the influence of a magnetic field. Transitions between these states involve changes in energy that can be facilitated by absorbing or emitting photons, highlighting how the protons' magnetic moments interact with the field.
Planck's Equation
Planck's equation is foundational in linking energy and frequency in photon-related processes. It is given by the formula:\[ \Delta E = h \cdot u \]where:
  • \( \Delta E \) is the change in energy.
  • \( h \) is Planck's constant, \( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \).
  • \( u \) is the frequency of the photon.
In this exercise, the frequency of 22.7 MHz is linked to the required energy change for transitions between proton spin states in the magnetic field. By rearranging Planck鈥檚 equation and using known constants such as the magnetic moment, we can derive the necessary magnetic field strength needed for this specific frequency. This demonstrates the interplay of quantum mechanical principles with physical phenomena.

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Most popular questions from this chapter

Consider the nuclear reaction \(^{4}_{2}He\) + \(^{7}_{3}Li\) \(\rightarrow\) X + \(^{1}_{0}n\) where X is a nuclide. (a) What are Z and A for the nuclide X? (b) Is energy absorbed or liberated? How much?

(a) If a chest x ray delivers 0.25 mSv to 5.0 kg of tissue, how many \(total\) joules of energy does this tissue receive? (b) Natural radiation and cosmic rays deliver about 0.10 mSv per year at sea level. Assuming an RBE of 1, how many rem and rads is this dose, and how many joules of energy does a 75-kg person receive in a year? (c) How many chest x rays like the one in part (a) would it take to deliver the same \(total\) amount of energy to a 75-kg person as she receives from natural radiation in a year at sea level, as described in part (b)?

Many radioactive decays occur within a sequence of decays for example, \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\). The half-life for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay is \(2.46 \times 10^{5}\) y, and the half-life for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay is \(7.54 \times 10^{4}\) y. Let 1 refer to \(^{234}_{92}U\), 2 to \(^{230}_{88}Th\), and 3 to \(^{226}_{84}Ra\); let \(\lambda\)1 be the decay constant for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay and \(\lambda\)2 be the decay constant for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay. The amount of \(^{230}_{88}Th\) present at any time depends on the rate at which it is produced by the decay of \(^{234}_{92}U\) and the rate by which it is depleted by its decay to \(^{226}_{84}Ra\). Therefore, \(d$$N_{2}\)(\(t\))/\(d$$t\) = \(\lambda$$_1$$N$$_1\)(\(t\)) - \(\lambda$$_2$$N$$_2\)(\(t\)). If we start with a sample that contains \(N_{10}\) nuclei of \(^{234}_{92}U\) and nothing else, then \(N{(t)}\) = \(N_{01}\)e\(^{-\lambda_{1}t}\). Thus \(dN_{2}(t)/dt\) = \(\lambda\)1\(N_{01}\)e\(^{-\lambda_{1}t}\)- \(\lambda2N_{2}(t)\). This differential equation for \(N_{2}(t)\) can be solved as follows. Assume a trial solution of the form \(N_{2}(t)\) = \(N_{10} [h_{1}e^{-\lambda_1t}\) + \(h_{2}e^{-\lambda_2t}\)] , where \(h_{1}\) and \(h_{2}\) are constants. (a) Since \(N_{2}(0)\) = 0, what must be the relationship between \(h_{1}\) and \(h_{2}\)? (b) Use the trial solution to calculate \(dN_{2}(t)/dt\), and substitute that into the differential equation for \(N_{2}(t)\). Collect the coefficients of \(e^{-\lambda_{1}t}\) and \(e^{-\lambda_{2}t}\). Since the equation must hold at all t, each of these coefficients must be zero. Use this requirement to solve for \(h_{1}\) and thereby complete the determination of \(N_{2}(t)\). (c) At time \(t = 0\), you have a pure sample containing 30.0 g of \(^{234}_{92}U\) and nothing else. What mass of \(^{230}_{88}Th\) is present at time \(t = 2.46 \times 10^{5}\) y, the half-life for the \(^{234}_{92}U\) decay?

The unstable isotope \(^4$$^0\)K is used for dating rock samples. Its half-life is 1.28 \(\times\) 10\(^9\) y. (a) How many decays occur per second in a sample containing 1.63 \(\times\) 10\(^-$$^6\) g of \(^4$$^0\)K? (b) What is the activity of the sample in curies?

Which reaction produces \(^{131}\)Te in the nuclear reactor? (a) \(^{130}Te + n \rightarrow ^{131}Te\); (b) \(^{130}I + n \rightarrow ^{131}Te\); (c) \(^{132}Te + n \rightarrow ^{131}Te\); (d) \(^{132}I + n \rightarrow ^{131}Te\).
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