How do I calculate Heisenburg's Uncertainty Principle?

• Ng Yau Ming
In summary: The answer to the question has a ±5% to the answer, so even if the answer is 1.9E-24, my answer of 1.8842808E-24 should be accepted. So that rules significant figures out.Yeah, I think that is probably it. Thanks for catching that! Yeah, I think that is probably it. Thanks for catching that!
Ng Yau Ming
Homework Statement
The uncertainty in the position of an electron along an x axis is given as 28 pm. What is the least uncertainty in any simultaneous measurement of the momentum component px of this electron?
Relevant Equations
(Uncertainty in x)*(Uncertainty in momentum in x) >= hbar/2
hbar=h/(2*pi)
I know Heisenburg's Uncertainty Principle states that there has to be a minimum amount of uncertainty. Where the minimum uncertainty is hbar/2.

My attempt at the solution

Uncertainty in x = 28E-12 m (Turn pm into m)
h=6.63E-34 (constant)
hbar=1.05519E-34 (constant)
hbar/2 = 5.275986363E-35 (constant)

Since the formula is
(Uncertainty in x)*(Uncertainty in momentum in x) >= hbar/2
and that it is asking for the minimum uncertainty, therefor minimum amount of uncertainty would be when it is hbar/2
(Uncertainty in x)*(Uncertainty in momentum in x) = hbar/2

So to calculate uncertainty in momentum
(Uncertainty in momentum in x)= (hbar/2)/(Uncertainty in x)

The answer I got is 1.884280844E-24
However this is wrong.
The only thing I can think of is that I got my magnitudes wrong? 28 pm should be 28E-12 m correct?
Im terribly sorry. I know this should be a simple question but I cannot figure out what is wrong with my calculations.

Assistance would be much appreciated.
Thank you.

Edit : Disclaimer I know this is on the homework forum but I need to tell you this homework is graded. I am sorry if I should not post this if it has to be graded

Ng Yau Ming said:
Homework Statement: The uncertainty in the position of an electron along an x-axis is given as 28 pm. What is the least uncertainty in any simultaneous measurement of the momentum component px of this electron?
Homework Equations: (Uncertainty in x)*(Uncertainty in momentum in x) >= hbar/2
hbar=h/(2*pi)

I know Heisenburg's Uncertainty Principle states that there has to be a minimum amount of uncertainty. Where the minimum uncertainty is hbar/2.

My attempt at the solution

Uncertainty in x = 28E-12 m (Turn pm into m)
h=6.63E-34 (constant)
hbar=1.05519E-34 (constant)
hbar/2 = 5.275986363E-35 (constant)

Since the formula is
(Uncertainty in x)*(Uncertainty in momentum in x) >= hbar/2
and that it is asking for the minimum uncertainty, therefor minimum amount of uncertainty would be when it is hbar/2
(Uncertainty in x)*(Uncertainty in momentum in x) = hbar/2

So to calculate uncertainty in momentum
(Uncertainty in momentum in x)= (hbar/2)/(Uncertainty in x)

The answer I got is 1.884280844E-24
However this is wrong.
The only thing I can think of is that I got my magnitudes wrong? 28 pm should be 28E-12 m correct?
Im terribly sorry. I know this should be a simple question but I cannot figure out what is wrong with my calculations.

Assistance would be much appreciated.
Thank you.

Edit : Disclaimer I know this is on the homework forum but I need to tell you this homework is graded. I am sorry if I should not post this if it has to be graded

Why do you think it is wrong?

Note that you have a lot of (unjustifiable) significant figures in your answer.

PeroK said:
Why do you think it is wrong?

Note that you have a lot of (unjustifiable) significant figures in your answer.

I don't quite understand what you mean by unjustifiable. Sorry english isn't my strong language. But Ill try to explain what I think is wrong

Most likely it is the significant figure section.
The answer wants it in Ns, so I changed pm to m. (28 pm to (28E-12 m or 2.8E-11 m))
Used the Js form of h instead of eVs (h = 6.63x10^34 Js)
So they are all in standard units and I am not using the wrong numbers
I tried to check whether i misunderstood pm as picometers, when its supposed to be something else but the only alternative was P for peta (10E15), which I doubt is the case.

I don't think I am misunderstanding it, as it tells us the minimum amount of uncertainty. So taking the lowest possible value, hbar/2 should be the minumum.

Ng Yau Ming said:
I don't quite understand what you mean by unjustifiable. Sorry english isn't my strong language. But Ill try to explain what I think is wrong

Most likely it is the significant figure section.
The answer wants it in Ns, so I changed pm to m. (28 pm to (28E-12 m or 2.8E-11 m))
Used the Js form of h instead of eVs (h = 6.63x10^34 Js)
So they are all in standard units and I am not using the wrong numbers
I tried to check whether i misunderstood pm as picometers, when its supposed to be something else but the only alternative was P for peta (10E15), which I doubt is the case.

I don't think I am misunderstanding it, as it tells us the minimum amount of uncertainty. So taking the lowest possible value, hbar/2 should be the minumum.
Look up significant figures. You only have two in the value for ##\Delta x##. So you are not justified in having more than two in your answer.

PeroK said:
Look up significant figures. You only have two in the value for ##\Delta x##. So you are not justified in having more than two in your answer.

The answer to the question has a ±5% to the answer, so even if the answer is 1.9E-24, my answer of 1.8842808E-24 should be accepted. So that rules significant figures out.

So maybe my next mistake is an error when conversion of pm to m?
Or maybe when calculating the Heisenburg Uncertainty principal I should be using the eV version instead of the Js?

Ng Yau Ming said:
The answer to the question has a ±5% to the answer, so even if the answer is 1.9E-24, my answer of 1.8842808E-24 should be accepted. So that rules significant figures out.
You are missing the point. You are stating a meaningless number in that the precision that you have used is totally unjustified. 1.88 would be barely justified. 1.8842808 is just silly.

And by the way, this is NOT just quibbling. If you don't understand it, you should think about it / study it until you do. If you don't understand significant digits, there are some problems where it will cause you grief.

Ng Yau Ming said:
The answer to the question has a ±5% to the answer, so even if the answer is 1.9E-24, my answer of 1.8842808E-24 should be accepted.

What do you mean by this? Accepted by what? You leave me with no idea why you think your answer is wrong.

You definitely shouldn't be converting to ##eV##.

PeroK said:
What do you mean by this? Accepted by what? You leave me with no idea why you think your answer is wrong.

You definitely shouldn't be converting to ##eV##.

I'm sorry. Its an online homework that accepts a range of ±5% of the actual value. And usually the actual value given has lots of decimal places. I put lots of decimal places in the answer just so I don't accidently fall out of the range. So any value I type as long as it is within the range will be accepted.

Also I mixed up significant figure for that (x10^n) thing here.

Ng Yau Ming said:
I don't quite understand what you mean by unjustifiable. Sorry english isn't my strong language. But Ill try to explain what I think is wrong

Most likely it is the significant figure section.
The answer wants it in Ns, so I changed pm to m. (28 pm to (28E-12 m or 2.8E-11 m))
Used the Js form of h instead of eVs (h = 6.63x10^34 Js)
So they are all in standard units and I am not using the wrong numbers
I tried to check whether i misunderstood pm as picometers, when its supposed to be something else but the only alternative was P for peta (10E15), which I doubt is the case.

I don't think I am misunderstanding it, as it tells us the minimum amount of uncertainty. So taking the lowest possible value, hbar/2 should be the minumum.

I am terribly sorry for that. What I probably mean is magnitude.

Edit : The question did not specify to use 2 significant figures
Edit 2 : Previously they did also do not mind the amount of significant figures

Last edited:
Ng Yau Ming said:
I'm sorry. Its an online homework that accepts a range of ±5% of the actual value. And usually the actual value given has lots of decimal places. I put lots of decimal places in the answer just so I don't accidently fall out of the range. So any value I type as long as it is within the range will be accepted.

Also I mixed up significant figure for that (x10^n) thing here.
I am terribly sorry for that. What I probably mean is magnitude.

Edit : The question did not specify to use 2 significant figures

I get what you get for SI units. Does the question specify the units for the answer?

PeroK said:
I get what you get for SI units. Does the question specify the units for the answer?

Yes over here
Ng Yau Ming said:
I don't quite understand what you mean by unjustifiable. Sorry english isn't my strong language. But Ill try to explain what I think is wrong

Most likely it is the significant figure section.
The answer wants it in Ns, so I changed pm to m. (28 pm to (28E-12 m or 2.8E-11 m))
Used the Js form of h instead of eVs (h = 6.63x10^34 Js)
So they are all in standard units and I am not using the wrong numbers
I tried to check whether i misunderstood pm as picometers, when its supposed to be something else but the only alternative was P for peta (10E15), which I doubt is the case.

I don't think I am misunderstanding it, as it tells us the minimum amount of uncertainty. So taking the lowest possible value, hbar/2 should be the minumum.

The answer wants it in Ns or kgm/s

Ng Yau Ming said:
Yes over hereThe answer wants it in Ns or kgm/s

I'd just move on. It's a conceptually simple question in any case.

Ng Yau Ming
To add, you can correct the uncertainty in the last name from Heisenburg into the (almost?) 100% certain Heisenberg. It will make the HW seem better ;). Not criticizing, you're not a native English speaker, though you speak English way better than I will likely speak your language.

PeroK said:
I'd just move on. It's a conceptually simple question in any case.

Understood. I'll have the homework coordinator look at it in a few days and focus on the others for the time being.
Thank you.

WWGD said:
To add, you can correct the uncertainty in the last name from Heisenburg into the (almost?) 100% certain Heisenberg. It will make the HW seem better ;). Not criticizing, you're not a native English speaker, though you speak English way better than I will likely speak your language.
Understood.
Thank you

WWGD
Textbooks occasionally have wrong answers in the back, so I wouldn’t be surprised if online homework systems have them also.

Your calculations look OK to me (and my calculator), except for the number of significant figures in the final answer. On a written homework or exam, I’d accept only two or three digits, two being the strictly correct number and three to be generous.

jtbell said:
Textbooks occasionally have wrong answers in the back, so I wouldn’t be surprised if online homework systems have them also.

Your calculations look OK to me (and my calculator), except for the number of significant figures in the final answer. On a written homework or exam, I’d accept only two or three digits, two being the strictly correct number and three to be generous.
Understood. I will keep that in mind during the examination. However in the exercises they will usually give out answers with a lot of digits and I usually put a lot of digits too just to ensure I don't fall out of that 5% range.

Ng Yau Ming said:
I usually put a lot of digits too just to ensure I don't fall out of that 5% range.
1.88 has an error range of +/- 1/2 out of 188, so about .5% (that's 1/10th of 5%). Using more significant digits can't possibly put you outside of the 5% range.

Again, since you don't seem to understand significant digits, I suggest that you study about them until you do.

1. What is Heisenburg's Uncertainty Principle?

Heisenburg's Uncertainty Principle is a fundamental concept in quantum mechanics that states that it is impossible to know both the position and momentum of a particle simultaneously.

2. How do I calculate the uncertainty in position and momentum?

The uncertainty in position (Δx) and momentum (Δp) can be calculated using the following formula: Δx*Δp ≥ h/4π, where h is Planck's constant (6.626 x 10^-34 joule seconds).

3. Can the uncertainty principle be applied to macroscopic objects?

No, the uncertainty principle is only applicable to particles on a quantum scale. The effects of the uncertainty principle are negligible at macroscopic levels.

4. How does the uncertainty principle relate to wave-particle duality?

The uncertainty principle is a result of the wave-like properties of particles on a quantum scale. It explains why particles cannot be described as having a definite position and momentum at the same time.

5. How has the uncertainty principle impacted our understanding of the physical world?

The uncertainty principle has revolutionized our understanding of the physical world by introducing the concept of indeterminacy and challenging the classical Newtonian view of a deterministic universe. It has also led to the development of groundbreaking technologies such as the electron microscope and laser technology.

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