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This is known as the Heisenberg uncertainty principle . It is impossible to measure position x size 12{x} {} and momentum p size 12{p} {} simultaneously with uncertainties Δ x size 12{Δx} {} and Δ p size 12{Δp} {} that multiply to be less than h / size 12{h/4π} {} . 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 Δ x = 0 . 0100 nm size 12{Δx=0 "." "0100"`"nm"} {} . Thus the smallest uncertainty in momentum Δ p size 12{Δp} {} can be calculated using Δ x Δ p h /4 π size 12{ΔxΔp>= h"/4"π} {} . Once the uncertainty in momentum Δ p size 12{Δp} {} is found, the uncertainty in velocity can be found from Δ p = m Δ v size 12{Δp=mΔv} {} .

Solution for (a)

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

Δ x Δ p = h . size 12{ΔxΔp = { {h} over {4π} } } {}

Solving for Δ p size 12{Δp} {} and substituting known values gives

Δ p = h Δ x = 6.63 × 10 –34 J s ( 1.00 × 10 –11 m ) = 5 . 28 × 10 –24 kg m/s . size 12{Δp = { {h} over {4π Δx} } = { {" 6" "." "63 " times " 10" rSup { size 8{"–34"} } " J " cdot " s"} over {4π \( 1 "." "00" times " 10" rSup { size 8{"–11"} } " m" \) } } =" 5" "." "28 " times " 10" rSup { size 8{"–24"} } " kg " cdot " m/s " "." } {}

Thus,

Δ p = 5 . 28 × 10 –24 kg m/s = m Δ v . size 12{Δp =" 5" "." "28 " times " 10" rSup { size 8{"–24"} } " kg " cdot " m/s "= mΔv} {}

Solving for Δ v size 12{Δv} {} and substituting the mass of an electron gives

Δ v = Δ p m = 5 . 28 × 10 –24 kg m/s 9 . 11 × 10 –31 kg = 5 . 79 × 10 6 m/s . size 12{Δv = { {Δp} over {m} } = { {5 "." "28 " times " 10" rSup { size 8{"–24"} } " kg " cdot " m/s"} over {9 "." "11 " times " 10" rSup { size 8{"–31"} } " kg"} } =" 5" "." "79 " times " 10" rSup { size 8{6} } " m/s" "." } {}

Solution for (b)

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

KE e = 1 2 mv 2 = 1 2 ( 9.11 × 10 –31 kg ) ( 5.79 × 10 6 m/s ) 2 = (1.53 × 10 –17 J) ( 1 eV 1.60 × 10 –19 J ) = 95.5 eV. alignl { stack { size 12{"KE" size 8{e} = { {1} over {2} } ital "mv" rSup { size 8{2} } } {} #=" 0" "." 5 \( 9 "." "11 " times " 10" rSup { size 8{"–31"} } " kg" \) \( 5 "." "79 " times " 10" rSup { size 8{6} } " m/s" \) rSup { size 8{2} } {} # =" 1" "." "53 " times " 10" rSup { size 8{"–17"} } " J " cdot { {"1eV"} over {1 "." "60 " times " 10" rSup { size 8{"–19"} } " J"} } =" 95" "." "5 eV" "." {}} } {}

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.

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Questions & Answers

Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
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?
Emmanuel Reply
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 Reply
Please help!
Emmanuel
please help find dy/dx 2x-y/x+y
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?
kwame Reply
yes what is it?
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
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
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
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
what's the answer then
Julius
great Mudang
Kossi
please Ana explain 4000 ?
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
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
isidor
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
BRIAN Reply
I think from kinematics
Nancy
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
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
solve the formula's please
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
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?
Hayne Reply
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir
Practice Key Terms 6

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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