Aamc 8 #10

[bajoneswadup]

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For a given magnitude of B1, the nucleus with the nonzero precession frequency will be which of the following:

A: 4/2He
B: 16/8O
C:19/9 F
D:208/82Pb

B1 is a magnetic field vector.
Answer:C

I got the right answer, but isn't 16/8O paramagnetic too? This question is just diamagnetic vs paramagnetic right? O has 2p4 so...confused a little

 

[pfaction]

I reasoned that it was an even number of electrons.

 

[MrNeuro]

For a given magnitude of B1, the nucleus with the nonzero precession frequency will be which of the following:

A: 4/2He
B: 16/8O
C:19/9 F
D:208/82Pb

B1 is a magnetic field vector.
Answer:C

I got the right answer, but isn't 16/8O paramagnetic too? This question is just diamagnetic vs paramagnetic right? O has 2p4 so...confused a little

keyword is nucleus got me for a second then i remembered NMR = NUCLEAR magnetic resonance imaging then remembered 1H and 13C work

you're dealing pairing of protons thus you want an uneven number of protons

also remember that its NOT electrons b/c the end product of an MRI is an image thats created by photons of light

E=mc^2 so an electron w/ a really small mass when undergoing an energy transtion from a parallel spin state (in line w/ the magnetic field) back down to the normal energy level its going to release a photon of light E= hf = hc/lambda

 

[typicalindian]

If I remember correctly, the passage even tells you "an unpaired electron (or proton, can't remember off the top of my head)". Since the atomic # = # of protons and # of electrons (Assuming it's in its neutral state) the first thing you look for is the element with an odd atomic number.

 

[pfaction]

It's a good thing my reading comprehension has not increased.

 

[typicalindian]

It's a good thing my reading comprehension has not increased.

Definitely pay attention to the passage. There have been several questions throughout the aamc's that CAN be answered through background knowledge but alot of the time the answer is literally in the passage.