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Chapter 40: Problem 74

The precession frequency of the protons in a laboratory NMR spectrometer is \(15.35850 \mathrm{MHz}\). The magnetic moment of the proton is \(1.410608 \cdot 10^{-26} \mathrm{~J} / \mathrm{T}\), while its spin angular momentum is \(0.5272863 \cdot 10^{-34} \mathrm{~J}\) s. Calculate the magnitude of the magnetic field in which the protons are immersed.

Short Answer

Expert verified
Question: Calculate the magnitude of the magnetic field in which the protons are immersed, given the precession frequency of 15.35850 MHz, the magnetic moment of the proton as 1.410608 x 10^(-26) J/T, and the spin angular momentum of the proton as 0.5272863 x 10^(-34) J*s. Answer: The magnitude of the magnetic field in which the protons are immersed is approximately 0.578 T.

Step by step solution

01

Write down the given values

We are given the following values: - Precession frequency, \(\nu = 15.35850 \mathrm{MHz}\) - Magnetic moment of the proton, \(\mu = 1.410608 \times 10^{-26} \mathrm{J/T}\) - Spin angular momentum of the proton, \(s = 0.5272863 \times 10^{-34} \mathrm{J\cdot s}\)
02

Convert the precession frequency to Hz

We need to convert the precession frequency from MHz to Hz: \(\nu = 15.35850 \mathrm{MHz} = 15.35850 \times 10^6 \mathrm{Hz}\)
03

Rearrange the Larmor equation to solve for the magnetic field

The Larmor equation is given by: \(\nu = \frac{\mu B}{s}\) We need to solve for \(B\), the magnetic field magnitude. Rearranging the Larmor equation, we get: \(B = \frac{\nu \cdot s}{\mu}\)
04

Plug in the given values and calculate the magnetic field magnitude

Now, we can plug in the given values to find the magnetic field magnitude: \(B = \frac{(15.35850 \times 10^6 \mathrm{Hz}) \cdot (0.5272863 \times 10^{-34} \mathrm{J\cdot s})}{1.410608 \times 10^{-26} \mathrm{J/T}}\) Computing the values, we get: \(B \approx 0.578 \mathrm{T}\)
05

State the final answer

The magnitude of the magnetic field in which the protons are immersed is approximately \(0.578 \mathrm{T}\).

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Most popular questions from this chapter

Consider the following fusion reaction, which allows stars to produce progressively heavier elements: \({ }_{2}^{3} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{4}^{7} \mathrm{Be}+\gamma\). The mass of \({ }_{2}^{3} \mathrm{He}\) is \(3.016029 \mathrm{u}\), the mass of \({ }_{2}^{4}\) He is \(4.002603 \mathrm{u}\), and the mass of \({ }_{4}^{7} \mathrm{Be}\) is \(7.0169298 \mathrm{u}\). The atomic mass unit is \(u=1.66 \cdot 10^{-27} \mathrm{~kg} .\) Assuming the Be atom is at rest after the reaction and neglecting any potential energy between the atoms and kinetic energy of the He nuclei, calculate the minimum possible energy and maximum possible wavelength of the photon \(\gamma\) that is released in this reaction.

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