# 29.7 Probability: the heisenberg uncertainty principle  (Page 3/11)

 Page 3 / 11

This is known as the Heisenberg uncertainty principle . It is impossible to measure position $x$ and momentum $p$ simultaneously with uncertainties $\Delta x$ and $\Delta p$ that multiply to be less than $h/4\pi$ . Neither uncertainty can be zero. Neither uncertainty can become small without the other becoming large. A small wavelength allows accurate position measurement, but it increases the momentum of the probe to the point that it further disturbs the momentum of a system being measured. For example, if an electron is scattered from an atom and has a wavelength small enough to detect the position of electrons in the atom, its momentum can knock the electrons from their orbits in a manner that loses information about their original motion. It is therefore impossible to follow an electron in its orbit around an atom. If you measure the electron’s position, you will find it in a definite location, but the atom will be disrupted. Repeated measurements on identical atoms will produce interesting probability distributions for electrons around the atom, but they will not produce motion information. The probability distributions are referred to as electron clouds or orbitals. The shapes of these orbitals are often shown in general chemistry texts and are discussed in The Wave Nature of Matter Causes Quantization .

## Heisenberg uncertainty principle in position and momentum for an atom

(a) If the position of an electron in an atom is measured to an accuracy of 0.0100 nm, what is the electron’s uncertainty in velocity? (b) If the electron has this velocity, what is its kinetic energy in eV?

Strategy

The uncertainty in position is the accuracy of the measurement, or $\Delta x=0\text{.}\text{0100}\phantom{\rule{0.25em}{0ex}}\text{nm}$ . Thus the smallest uncertainty in momentum $\Delta p$ can be calculated using $\Delta x\Delta p\ge h\text{/4}\pi$ . Once the uncertainty in momentum $\Delta p$ is found, the uncertainty in velocity can be found from $\Delta p=m\Delta v$ .

Solution for (a)

Using the equals sign in the uncertainty principle to express the minimum uncertainty, we have

$\Delta x\Delta p=\frac{h}{4\pi }.$

Solving for $\Delta p$ and substituting known values gives

$\Delta p=\frac{h}{4\pi \Delta x}=\frac{\text{6.63}×{\text{10}}^{\text{–34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot \text{s}}{4\pi \left(1.00×{\text{10}}^{\text{–11}}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}=\text{5}\text{.}\text{28}×{\text{10}}^{\text{–24}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}\text{.}$

Thus,

$\Delta p=\text{5}\text{.}\text{28}×{\text{10}}^{\text{–24}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}=m\Delta v.$

Solving for $\Delta v$ and substituting the mass of an electron gives

$\Delta v=\frac{\Delta p}{m}=\frac{5\text{.}\text{28}×{\text{10}}^{\text{–24}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}}{9\text{.}\text{11}×{\text{10}}^{\text{–31}}\phantom{\rule{0.25em}{0ex}}\text{kg}}=\text{5}\text{.}\text{79}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m/s}\text{.}$

Solution for (b)

Although large, this velocity is not highly relativistic, and so the electron’s kinetic energy is

$\begin{array}{lll}{\text{KE}}_{e}& =& \frac{1}{2}{\text{mv}}^{2}\\ & =& \frac{1}{2}\left(\text{9.11}×{\text{10}}^{\text{–31}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(\text{5.79}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}\\ & =& \text{(1.53}×{\text{10}}^{\text{–17}}\phantom{\rule{0.25em}{0ex}}\text{J)}\left(\frac{\text{1 eV}}{\text{1.60}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J}}\right)=\text{95.5 eV.}\end{array}$

Discussion

Since atoms are roughly 0.1 nm in size, knowing the position of an electron to 0.0100 nm localizes it reasonably well inside the atom. This would be like being able to see details one-tenth the size of the atom. But the consequent uncertainty in velocity is large. You certainly could not follow it very well if its velocity is so uncertain. To get a further idea of how large the uncertainty in velocity is, we assumed the velocity of the electron was equal to its uncertainty and found this gave a kinetic energy of 95.5 eV. This is significantly greater than the typical energy difference between levels in atoms (see [link] ), so that it is impossible to get a meaningful energy for the electron if we know its position even moderately well.

Why is there no 2nd harmonic in the classical electron orbit?
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what kind of error do you think? and work is held by which force?
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I am now in this group
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so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
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LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
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Nancy
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Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
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mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
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Julius
great Mudang
Kossi
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
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its not possible
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í want the working procedure
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mohammed
physics is the science that studies the non living nature
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isidor
what is phyacs
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I think from kinematics
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You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
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Ana
benjamin
those are the three .. what you wanna solve ?
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For first equation simply integrate formula of acceleration in the limit v and u
Tripti
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similarly for 3 one integrate acceleration again by multiplying and dividing term ds
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Ana
what is the question dear
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find the distance
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Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir