AAMC 6 physics mass spec

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chiddler

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Lithium ions are shot in an Efield and the time it takes to traverse the field is measured. Peak 3 = 6-Li cation, peak 4 = 7-Li cation. So the first ones must be the corresponding double cations.

96Ai5.png


Question:

Peak 5 in figure 2 originates from a different atomic species. Given where the peak appears, and assuming that it corresponds to a singly ionized atom, we can say that the atoms of this species probably have:

A. More protons and more neutrons than 7-Li

What is the proper way to reason this question?

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I mean, the heavier it is, the larger the radius is, the longer it will take? I dunno, I tried to do r = mv/qb.
 
eh. stupid of me. should of just written all the answers for constrast, sorry.

B was like more protons fewer neutrons
C fewer protons more neutrons
D fewer fewer.

Ok here's an idea. If more protons and fewer neutrons, or vice versa, the mass stays about the same and therefore we would see a similar result as one of the other lithium ions.

Since it is very saliently heavier and slower, it must therefore hold both more protons and more neutrons.

I can think of two rebuttles to this though. Too lazy to type out. Just an idea...
 
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Peak 5 ions take a longer time to reach the anode so they are heavier. Heavier means more protons and more neutrons. Just more protons or just more neutrons would mean heavier as well but once you start doing fewer of either one you will be reducing the mass. So I think safest answer is to "load" up on both protons and neutrons (A).
 
eh. stupid of me. should of just written all the answers for constrast, sorry.

B was like more protons fewer neutrons
C fewer protons more neutrons
D fewer fewer.

Ok here's an idea. If more protons and fewer neutrons, or vice versa, the mass stays about the same and therefore we would see a similar result as one of the other lithium ions.

Since it is very saliently heavier and slower, it must therefore hold both more protons and more neutrons.

I can think of two rebuttles to this though. Too lazy to type out. Just an idea...

i think what i've written below may answer your question

Peak 5 ions take a longer time to reach the anode so they are heavier. Heavier means more protons and more neutrons. Just more protons or just more neutrons would mean heavier as well but once you start doing fewer of either one you will be reducing the mass. So I think safest answer is to "load" up on both protons and neutrons (A).

i actually don't think thats precisely why thats the correct answer i think it has more to do w/ proportionality and the equation

we know the following

first we know a larger radius means a longer time till collision so were looking for what will produce the largest value for radius

from what pfaction wrote
r= mv/qB

v and B for all purposes are constant

for our purposes neutrons and protons (nucleons) have the same mass so mass of 1 n = mass of 1 p

however of those two nucleon species one also has another affect on our equation q (charge) so from the choices we have

a) increasing protons and neutrons
this would yield a ratio of 2:1 for m and q respectively (2 nucleons = 2 amu added and 1 C of charge)
b) increasing protons and fewer neutrons yields a 1:1 ratio
c) increasing neutrons fewer protons 1:0(meaning nothing added)
d) fewer of both means 0:0

i think a is the best answer this question is loaded tho b/c i dunno if youre adding and possibly removing or just adding

weird ass question

wheres milski? loool
 
mrneuro,

kuntash repeated exactly what i wrote. and like i said, i can think of two reasons that this is wrong. except i don't remember the second reason and can only think of one.

1. increasing protons decreasing neutrons does not necessarily indicate that they are equal in number. meaning we might add 3 protons but only take away 1 neutron. it is not very clear. thus, this option (and its counterpart increase neutrons decrease protons) is capable of a net increase in mass which makes it potentially valid just as A is.

In response to what you wrote, "corresponds to a singly ionized atom" is in the question. Therefore we assume that charge does not vary. In other words, q is constant.
 
mrneuro,

kuntash repeated exactly what i wrote. and like i said, i can think of two reasons that this is wrong. except i don't remember the second reason and can only think of one.

1. increasing protons decreasing neutrons does not necessarily indicate that they are equal in number. meaning we might add 3 protons but only take away 1 neutron. it is not very clear. thus, this option (and its counterpart increase neutrons decrease protons) is capable of a net increase in mass which makes it potentially valid just as A is.

In response to what you wrote, "corresponds to a singly ionized atom" is in the question. Therefore we assume that charge does not vary. In other words, q is constant.

ok well then yeah load up on mass like kuntash said
 
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