Acceleration of Uranium 238 ion through a potential difference

In summary: I'll pass on that. Get on with your homework, is my advice.Are you just thinking that I take absolute value...or did you mean something else?I'll pass on that. Get on with your homework, is my advice.
  • #1
Nway
34
0
Homework Statement
Below.
Relevant Equations
1/2mv^2
I don't understand why the Uranium 238 ions are accelerated
1674979863035.png

I think ##\Delta V = -2000 V## to accelerate since the ion would be accelerated by more postive charges so ## V_i > V_f ##
 

Attachments

  • 1674979830002.png
    1674979830002.png
    13.1 KB · Views: 64
Physics news on Phys.org
  • #2
I don't understand your question.
 
  • Like
Likes Nway
  • #3
PeroK said:
I don't understand your question.
I don't understand why ##\Delta V = 2000 V ## in the solutions.
 
  • #4
Nway said:
I don't understand why ##\Delta V = 2000 V ## in the solutions.
What do you think it should be?
 
  • Like
Likes Nway
  • #5
PeroK said:
What do you think it should be?
Because ##\Delta V = V_f - V_i ## so if ##\Delta V = 2000 V ## Then ##V_f = 2000V ## which only accelerates if q is negative. But q postive in this question sir.
 
  • #6
Nway said:
Because ##\Delta V = V_f - V_i ## so if ##\Delta V = 2000 V ## Then ##V_f = 2000V ## which only accelerates if q is negative.
Perhaps assume that the aparatus is configured so that the Uranium ion is accelerated.
 
  • Like
Likes Nway and Redbelly98
  • #7
PeroK said:
Perhaps assume that the aparatus is configured so that the Uranium ion is accelerated.
How would this work?
 
  • #8
Nway said:
How would this work?
You start the Uranium ion at the appropriate side of the potential difference. Like, you might start a ball at the top of a hill rather than the bottom.
 
  • Like
Likes Nway
  • #9
PeroK said:
You start the Uranium ion at the appropriate side of the potential difference. Like, you might start a ball at the top of a hill rather than the bottom.
##q\Delta V = 0.5mv^2 ##
##v = (\frac {2q\Delta V}{m})^{1/2} ##

However, ##\Delta V = -2000 V## since ##V_i = 2000## (more postive charges so will accelerate the ##U-238## cation) and ##V_f = 0 ##

However since q is positive then this would give a imaginary velocity as there is a negative in numerator. I don't understand why solutions say ##\Delta V = 2000 V##. I think ##\Delta V = -2000 V##

See what I mean?
 
  • #10
Nway said:
##q\Delta V = 0.5mv^2 ##
##v = (\frac {2q\Delta V}{m})^{1/2} ##

However, ##\Delta V = -2000 V## since ##V_i = 2000## (more postive charges so will accelerate the ##U-238## cation) and ##V_f = 0 ##

However since q is positive then this would give a imaginary velocity as there is a negative in numerator. I don't understand why solutions say ##\Delta V = 2000 V##. I think ##\Delta V = -2000 V##

See what I mean?
A difference is a magnitude. It doesn't have a sign. The difference between 2 and 3 is the same as the difference between 3 and 2.
 
  • Like
Likes Nway
  • #11
haruspex said:
A difference is a magnitude. It doesn't have a sign. The difference between 2 and 3 is the same as the difference between 3 and 2.
Why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs. Isn't ##\Delta V= V_f - V_i## ?
 
  • #12
Nway said:
See what I mean?
If an aparatus has a potential difference of ##2000 \ V##, then positive charges and negative charges may both be accelerated by it (albeit in opposite directions).
 
  • Like
Likes Nway
  • #13
PeroK said:
If an aparatus has a potential difference of ##2000 \ V##, then positive charges and negative charges may both be accelerated by it (albeit in opposite directions).
But why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs?
 
  • #14
Nway said:
But why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs?
I would say it does and specifying ##-2000 \ V## in the question would be more precise. But, it's clear what the question intends.
 
  • Like
Likes Nway
  • #15
PeroK said:
I would say it does and specifying ##-2000 \ V## in the question would be more precise. But, it's clear what the question intends.
But how can ## - 2000 V## be true since then the velocity will be purely imaginary?

Plugging in numbers:

##v = (\frac {-4000q}{m})^{1/2} ## but q and m is positive so ##v = i(\frac {4000q}{m})^{1/2} ##
 
  • Sad
Likes PeroK
  • #16
Nway said:
But how can ## - 2000 V## be true since then the velocity will be purely imaginary?

Plugging in numbers:

##v = (\frac {-4000q}{m})^{1/2} ## but q and m is positive so ##v = i(\frac {4000q}{m})^{1/2} ##
That's completely wrong. If you want to say "the Uranium ion is not accelerated, because the potential difference has the wrong sign", then that is your answer and you can move on to the next question.
 
  • Like
Likes Nway
  • #17
PeroK said:
That's completely wrong. If you want to say "the Uranium ion is not accelerated, because the potential difference has the wrong sign", then that is your answer and you can move on to the next question.
Sorry sir, could you explain a bit more?

Didn't you say that potential difference dose have a sign? And that sign is ##\Delta V = -2000V##?

But @haruspex disagreed.
 
  • #18
Nway said:
Sorry sir, could you explain a bit more?

Didn't you say that potential difference dose have a sign? And that sign is ##\Delta V = -2000V##?

But @haruspex disagreed.
I'll pass on that. Get on with your homework, is my advice.
 
  • Like
  • Care
Likes nasu and Nway
  • #19
PeroK said:
I'll pass on that. Get on with your homework, is my advice.
Are you just thinking that I take absolute value of the p.d? So then ##|\Delta V = - 2000V| = 2000 V ##

However, it seems to me like black magic to take the absolute value to get rid of the - sign and the imaginary velocity.
 
  • #20
The question doesn't specify the sign of the potential difference at all. However, if the positive uranium ion started next to the negative electrode, it would stick to it, and not be accelerated, whereas if it started next to the positive electrode, it would be repelled by it and accelerated towards the negative electrode. Since the question does specify that the ion is accelerated, we can conclude that it started next to the positive electrode.
You can also ignore the sign of the potential difference, and compute the amount of kinetic energy it gained from it, wich must be positive, since the ion was accelerated.
 
  • Like
Likes Redbelly98, Nway and nasu
  • #21
Nway said:
Why dose voltage difference not have a sign when ##V_f## and ##V_i## have signs. Isn't ##\Delta V= V_f - V_i## ?
Because that is what "difference" means. It does not specify which way to do the subtraction. What is the difference in height between Everest and Nanga Parbat? Would you answer with a negative value just because I put them in that order?
The question says the ions are accelerated "through a potential difference". It does not say "up through" or "down through", so still does not imply a direction.
 
  • Like
Likes PeroK and Nway
  • #22
haruspex said:
Because that is what "difference" means. It does not specify which way to do the subtraction. What is the difference in height between Everest and Nanga Parbat? Would you answer with a negative value just because I put them in that order?
The question says the ions are accelerated "through a potential difference". It does not say "up through" or "down through", so still does not imply a direction.
Oh I though difference meant finial - inital. But I guess initial - finial would also be ok if you took the absolute value.
 
  • #23
haruspex said:
Because that is what "difference" means. It does not specify which way to do the subtraction. What is the difference in height between Everest and Nanga Parbat? Would you answer with a negative value just because I put them in that order?
The question says the ions are accelerated "through a potential difference". It does not say "up through" or "down through", so still does not imply a direction.
p.d is a scalar thought, but would a negative sign sort of give it a direction?
 
  • #24
Nway said:
I thought difference meant final - initial.
I would use "change" for that.
Nway said:
p.d is a scalar thought, but would a negative sign sort of give it a direction?
Height is also a scalar, and scalars have direction.
 
  • Like
Likes Nway
  • #25
haruspex said:
I would use "change" for that.

Height is also a scalar, and scalars have direction.
Ain't the difference between scalars and vectors is that scalars have only two possible discrete directions while vectors have COUNTIOUS range of directions?

They normally teach us in school that scalars have magnitude and vectors have magnitude and direction.
 
  • #26
Nway said:
Ain't the difference between scalars and vectors is that scalars have only two possible discrete directions while vectors have COUNTIOUS range of directions?
Unless it's a 1D vector space. Some say scalars do form a 1D vector space; that's true except that you have to throw away some structure. E.g. you cannot divide by a vector.
Nway said:
They normally teach us in school that scalars have magnitude and vectors have magnitude and direction.
A common blunder. Scalars, as you note, do have direction, except that, like a 1D vector, it is only a choice of two directions.
 
  • Like
Likes Nway
  • #27
haruspex said:
A common blunder. Scalars, as you note, do have direction, except that, like a 1D vector, it is only a choice of two directions.
I prefer to think that "direction" is a vector attribute only and that scalars have a sign, positive or negative, but not a direction. When I see ##v_x=-3~##m/s, my interpretation is that the x-component of the velocity vector, a scalar, is negative, in whatever way "negative" on the x-axis is defined. In 1D, where there is no need for axes labels, subscript ##x## is commonly dropped but the interpretation remains the same.

I agree about the common blunder. I suspect the confusion arises because of the dropped subscript and the blurring of what is meant by the result. In 1D, the equation ##v=-3~##m/s, can have only one interpretation formally written as ##\mathbf{v}=3~(-\mathbf{\hat x})~##m/s. Thus, ##v=-3~##m/s is meant to be a vector equation but the unit vector is omitted while the negative sign is retained.

If, as you say, scalars have direction, how would you write the scalar in this case to indicate the direction it has? How can you write a scalar equation when your intention is to write a vector equation? If you write ##v=-3~##m/s, then I think that would feed into the confusion. One would interpret ##3~##m/s as the scalar part and the negative sign as the direction part reinforcing the misconception that scalars are always positive. If you write ##v_x=-3~##m/s, then there is no direction involved in ##v_x## because the vector equation is ##\mathbf{v}=v_x~\mathbf{\hat x}## and that's where the direction is.
 
  • Like
Likes Nway
  • #28
kuruman said:
I prefer to think that "direction" is a vector attribute only and that scalars have a sign, positive or negative, but not a direction. When I see ##v_x=-3~##m/s, my interpretation is that the x-component of the velocity vector, a scalar, is negative, in whatever way "negative" on the x-axis is defined. In 1D, where there is no need for axes labels, subscript ##x## is commonly dropped but the interpretation remains the same.
We are so used to thinking of vectors in Cartesian coordinates. But go back to the statement that vectors have "magnitude and direction". That fits more naturally with polar coordinates. In one dimension, the direction consists of the sign.
kuruman said:
I suspect the confusion arises because of the dropped subscript and the blurring of what is meant by the result. In 1D, the equation ##v=-3~##m/s, can have only one interpretation formally written as ##\mathbf{v}=3~(-\mathbf{\hat x})~##m/s.
That's as magnitude x direction, but as scalar x 1D vector it's ##\mathbf{v}=-3~(\mathbf{\hat x})~##m/s.
kuruman said:
One would interpret 3 m/s as the scalar part and the negative sign as the direction part
No, you should interpret it as either magnitude 3, direction -, or as scalar -3 on whatever oriented axis has been specified (like, 'up').
 
  • Like
Likes Nway

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
271
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
2
Replies
55
Views
2K
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
767
  • Introductory Physics Homework Help
Replies
1
Views
815
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
265
Back
Top