Particle movement in field (TBR CBT 3 PS Number 25)

[ridethecliche]

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Question asks what happens to r when m is doubled. Explanation below.

The radius of the path followed by an ionized, radially accelerated, moving particle in a mass spectrometer's magnetic field is given by Equation 1:

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The velocity of the particle depends on its mass (see Equation 2). Substituting Equation 2 into Equation 1 gives Equation 3:

​

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The radius is proportional to the square root of the particle's mass (r ∝ (m)½). If the mass were doubled, the radius would increase by a factor of , which is answer choice B. The reason choice C is not correct is that not all particles enter the mass spectrometer's magnetic field with the same initial velocity. The lightest particles generally enter the field with the greatest velocity. If the velocities of all particles as they entered the field were equal, then the radius of the path of any particle would double uniformly along with its mass.

In eq 3, I think it's supposed to be r=

but anyway, the question asks what happens to r if you double m.
What happens with the m outside the square root. Also why is it (2)^ 1/2
as there's a denominator of m inside the root as well.

Thanks for the help.

 

[Rabolisk]

What is 2 * sqrt(1/2)?

 

[ilovemcat]

Yep, you're right.

Anyways, for this problem, mass is proportional to the radius.

Holding all variables constant (velocity, charge, and magnetic field), but doubling mass would increase the radius by a factor of 2. [Equation 1]
Mass however is inversely related to velocity. Doubling mass would also decrease velocity by a factor of 1/sqrt(2). [Equation 2]

So two things have clearly changed: mass and velocity. Both of these variables are proportional to radius.
Together, increasing radius by a factor of 2, but decreasing velocity by a factor of 1/sqrt(2), results in an overall change in radius by a factor of:

2/sqrt(2)

If this isn't listed as an answer choice, then I'd try solving it out further: 2 divided 1.4 (1.4 is the sqrt of 2 which is approx. 3/2) ---> 2 / (3/2) = 4/3 = 1.33

When you see 2^1/2 it just means the same thing as sqrt of. Similarily, 2^1/3 would mean, cuberoot of.

Anyways, 1.33 is approx. equal to sqrt(2) ~~ 1.4


There's another way to simplify this but I'm not really experienced with simplifying square roots, so I'll leave that to someone else. The square root of 2 is good to know though as it appears a lot in several problems.

 

[ridethecliche]

What is 2 * sqrt(1/2)?

It's root 2.

I see that root 2 is the multiplier for m out front but what about the m in the denominator? Doesn't that change anything? If not, why not?

 

[ilovemcat]

It's root 2.

But there's an m in the denominator and one in front as a multiplier.

Doesn't that change anything? If not, why not?

If it was a simple fraction you're multiplying against, the two m's would just cancel out and the radius wouldn't change. But that isn't the case here, which is why you have to consider m/sqrt(m).

"m" and the sqrt of "m" are two different values.

 

[Rabolisk]

It's root 2.

I see that root 2 is the multiplier for m out front but what about the m in the denominator? Doesn't that change anything? If not, why not?

No, 2 is the multiplier for m out front. The second part (sqrt(1/m) or 1/sqrt(m)) gives you 1/sqrt(2). Multiply together to get root 2.

 

[BHaus9]

Yeah, to solve problems like this having to do with relative magnitudes, just plug in the multiplicative value wherever you see the variable in question and then ignore all variables and values in the equation UNLESS the value is an operator (i.e. exponent).

In this case, plug in the value "2" wherever you see variable "m", and then replace all other figures with the value 1. Then, according to Eq 3, we have r = (2/(1*1)) * (1*1*1/2)^0.5
Simplified, r = 2 * (1/2)^0.5 = 2 / 2^0.5
By definition, anything divided by its own square root is equal to the square root, so r = 2^0.5

 

[ridethecliche]

Okay. Lesson learned. Do not try to do basic math past midnight.

Looks like I was forgetting basic square root rules.

I.e. (1/2)^.5
is the same as
1/ (2^.5)

Seriously brain get with it!

Thanks for humoring me. I'm a little ashamed/embarrassed right now.

 

[ridethecliche]

okay, i swear I'm not crazy, but if you substitute everything with 1 then don't you end up with 2/2 inside the square root?

 

[ridethecliche]

Yeah, to solve problems like this having to do with relative magnitudes, just plug in the multiplicative value wherever you see the variable in question and then ignore all variables and values in the equation UNLESS the value is an operator (i.e. exponent).

In this case, plug in the value "2" wherever you see variable "m", and then replace all other figures with the value 1. Then, according to Eq 3, we have r = (2/(1*1)) * (1*1*1/2)^0.5
Simplified, r = 2 * (1/2)^0.5 = 2 / 2^0.5
By definition, anything divided by its own square root is equal to the square root, so r = 2^0.5

Looking back on it, I think you made a mistake. The second half of the equation has a 2 in the numerator under the square root so the red bold 1 above should be 2.

Right?

 

[ilovemcat]

When you're considering how something will change, (ie. will it increase by a factor of 2, decrease by a factor of 2, etc.), you consider only the variable you're changing and ignore the rest. It turned out for this problem that changing one variable (mass) indirectly affected another variable (velocity), which is why we also had to consider how velocity changed. But consider a scenario where only one variable changes:

Let's use the equation for this problem as an example:

r = mv/qB

First let's consider the variables as are (unchanged). Then whatever mv/qB equals is what "r" equals.

However, let's say you instead doubled the magnetic field, while holding all the other variables constant.
Since the variables m,v, and q aren't changing, what we can say is that the "mv/q" portion of the equation equals some constant value.
This "constant" (mv/q) times 1/B will equal the new value of r.

Increasing B by a factor of 2, will decrease "r" by a factor of: 1/B ---> 1/2 x constant
Increasing B by a factor of 4, will decrease "r" by a factor of: 1/B ---> 1/4 x constant.
Increasing B by a factor of 8, will decrease "r" by a factor of: 1/B ---> 1/8 x constant.

As long as you don't change the other variables, they'll always equal some constant value.

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In the case of the last equation (equation 3), we see that the variables q and V aren't changing.
Therefore, just as I explained above the unchanged values would equal some constant, which we tend to ignore for simplicity.

Equation 3, would simplify to: m/sqrt(m) x constant
The constant here would equal: 1/qB x sqrt(2qV) ... which we just ignore because we're only trying to figure out how the overall value of "r" changed in comparison to it's original value (ie. increasing by some factor, or decreasing by some factor).

Does this make sense? Hopefully it helped.

 

[ridethecliche]

Yeah, to solve problems like this having to do with relative magnitudes, just plug in the multiplicative value wherever you see the variable in question and then ignore all variables and values in the equation UNLESS the value is an operator (i.e. exponent).

In this case, plug in the value "2" wherever you see variable "m", and then replace all other figures with the value 1. Then, according to Eq 3, we have r = (2/(1*1)) * (1*1*1/2)^0.5
Simplified, r = 2 * (1/2)^0.5 = 2 / 2^0.5
By definition, anything divided by its own square root is equal to the square root, so r = 2^0.5

I think I get it. You're just going to consider the things you care about and ignore all the other factors.

Will do!

 

[ilovemcat]

I think I get it. You're just going to consider the things you care about and ignore all the other factors.

Will do!

I feel like you're avoiding me for some reason, lol. But glad you atleast understand it.

 

[MedChallenge]

Same passage #27

Why can't III be an alternative? If we changed the position of the double filter to place it above the detector, wouldn't that work?

27. Which of the following changes would have to be made to the mass spectroscopy apparatus as it is described in the passage, if the cation source in the experiment were replaced with an anion source? I. The polarity of the battery would have to be reversed.
II. The orientation of the magnetic field would have to be reversed.
III. The location of the double filter would have to be changed.

A. I only
B. II only
C. I and II only
C is the best answer. In order to accelerate an anion rather than a cation from left to right, we must first change the polarity of the battery from the layout shown in Figure 1 (i.e., from a design used to accelerate the positively charged species). Remember, anions are accelerated to regions of higher voltage (greater positive charge), while cations are accelerated to regions of lower voltage (greater negative charge). If the magnetic field remained unchanged, the anions would be deflected to the bottom of the page. The magnetic field must be reversed to deflect the anions to the top of the page. The path of the anions should be the same as the path of the cations from the source, so the double filter should remain in place where it is in Figure 1. This reasoning makes both options I and II valid, so the best answer is choice C. The best answer is C.
D. II and III only

 

[ilovemcat]

Same passage #27

Why can't III be an alternative? If we changed the position of the double filter to place it above the detector, wouldn't that work?

27. Which of the following changes would have to be made to the mass spectroscopy apparatus as it is described in the passage, if the cation source in the experiment were replaced with an anion source? I. The polarity of the battery would have to be reversed.
II. The orientation of the magnetic field would have to be reversed.
III. The location of the double filter would have to be changed.

A. I only
B. II only
C. I and II only
C is the best answer. In order to accelerate an anion rather than a cation from left to right, we must first change the polarity of the battery from the layout shown in Figure 1 (i.e., from a design used to accelerate the positively charged species). Remember, anions are accelerated to regions of higher voltage (greater positive charge), while cations are accelerated to regions of lower voltage (greater negative charge). If the magnetic field remained unchanged, the anions would be deflected to the bottom of the page. The magnetic field must be reversed to deflect the anions to the top of the page. The path of the anions should be the same as the path of the cations from the source, so the double filter should remain in place where it is in Figure 1. This reasoning makes both options I and II valid, so the best answer is choice C. The best answer is C.
D. II and III only

I responded to this a while back:
http://forums.studentdoctor.net/showpost.php?p=10700522&postcount=5

 

[MedChallenge]

I responded to this a while back:
http://forums.studentdoctor.net/showpost.php?p=10700522&postcount=5

Sorry I didn't see it before, but thanks...you give good explanations👍