Hydrogen atoms are placed in an external magnetic field. The prot...

Question

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?

Accepted Answer

Hi everyone in this practice problem, we're being asked to determine the magnitude of the magnetic field. We'll have an N M R experiment where a physicist placing a sample of protons in a magnetic field where the protons are in a low energy spin up state, each of the protons can absorb a photon and undergo spin flipping to the higher energy spin down state. If the photons energy matches the energy difference between the two states for the N M R experiment to work, we're being asked to determine the magnitude of the magnetic field that will guarantee a spin flip between the energy levels. If the photons have a frequency of 64 megahertz, the options given are a 0.5 tesla b, 1.0 tesla, C 1.5 tesla and D 2.0 tesla. So the energy of a proton in a state parallel or spin up to the magnetic field is going to be equal to spin up U equals to negative mu Z multiplied by B. On the other hand, the spin down or the energy of a proton in a state anti parallel or spin down to the magnetic field is going to be equal to U equals to mu Z multiplied by B. Therefore, the energy difference or delta E between the two states parallel is going to be equals to two muz multiplied by B. This energy difference should be equal to the four times energy where delta E will be equal to two mu Z multiplied by B and will be equal to H multiplied by F. Therefore, looking into this equation right here where two mu Z B equals to H F, we can rec uh arrange this equation to get an equation for the magnetic field magnitude or B which will be equals to H F divided by two mu Z U Z is going to be equal to 2.7928 UN. And we can actually uh substitute o of the variable into the situation for the magnetic field or B into our H F divided by two mu Z. So let's start doing that. H is just the planks constant which is a constant which is 6.6 to 6 times 10 to the power of negative 34 2 seconds. F is the frequency given which is 64 megahertz. So 64 times 10 to the power of six Hertz divide that with two muz which is going to be two multiplied by 2.7928 un, 2.7928 Un UN is going to be 5.51 times 10 to the power of negative 27 per tesla, which is just going to be a constant. Calculating this, we will get the magnitude of the magnetic field B to then be equal to 1.5 tesla. So the magnetic field magnitude is going to be equal to 1.5 tesla which will correspond to option C in our answer choices. So option C will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this video. Thank you.

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