<< Chapter < Page Chapter >> Page >

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?


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 " "." } {}


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" "." {}} } {}


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.

Got questions? Get instant answers now!

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?
I am now in this group
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?
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
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
i dont think dere is any potential energy... by d virtue of no height present
there is compressed energy,dats only potential energy na?
yes.. but... how will u approach that question without The Height in the question?
Can you explain how you get 54J?
Because mine is 36J
got 36J too
OK the answer is 54J Babar is correct
Conservation of Momentum
woow i see.. can you give the formula for this
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!
please help find dy/dx 2x-y/x+y
By using the Quotient Rule dy/dx = 3y/(x +y)²
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
i think i m correct
But how?
what's the big bang?
kwame Reply
yes what is it?
it is the explanation of how the universe began
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
thanks guys
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
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
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
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?.
pls í need help
4000 work is done
speed=distance /time distance=speed/time
now use this formula
what's the answer then
great Mudang
please Ana explain 4000 ?
hey mudang there is a product of force and acceleration not force and displacement
@Mohammed answer is 0.8hours or 48mins
its not possible
í want the working procedure
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
I think from kinematics
You can find physics books at the library or online. That's how I started.
And yes, kinematics is usually where you can begin.
study basic algebra and calculus and can start from classical mechanics
yes think so but dimension is the best starting point
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
solve the formula's please
those are the three .. what you wanna solve ?
For first equation simply integrate formula of acceleration in the limit v and u
For second itegrate velocity formula by ising first equation
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
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
what is the question dear
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
find the distance
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?
I think the formulae used for this question is lambda=(ax)/D
Practice Key Terms 6

Get the best College physics course in your pocket!

Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?