<< Chapter < Page Chapter >> Page >
This figure has two parts. Part a shows a diagram of an electron source emitting waves that pass through two narrow slits in a barrier. A wave interference pattern results on the opposite side of the barrier that result in evenly spaced, relatively wide horizontal grey lines on a black surface. The black surface is placed at a distance beyond the barrier. Part b shows an electron source at the far left. Two gold arrows point toward two narrow horizontal slits in a barrier. Around these arrows are many small golden dots. On the right side of the barrier, the gold dots are more widely dispersed. A black surface a distance beyond the barrier shows evenly, yet widely dispersed gold dots. An arrow is present below this black surface and is labeled, “Time.” A second black surface is shown to the right that has many more golden dots that appear to be organizing into a pattern of evenly spaced horizontal line segments. A third black surface is shown even further to the right in which many more gold dots are shown in a very clearly established, evenly spaced pattern of horizontal line segments.
(a) The interference pattern for electrons passing through very closely spaced slits demonstrates that quantum particles such as electrons can exhibit wavelike behavior. (b) The experimental results illustrated here demonstrate the wave–particle duality in electrons. The electrons pass through very closely spaced slits, forming an interference pattern, with increasing numbers of electrons being recorded from the left image to the right. With only a few electrons recorded, it is clear that the electrons arrive as individual localized “particles,” but in a seemingly random pattern. As more electrons arrive, a wavelike interference pattern begins to emerge. Note that the probability of the final electron location is still governed by the wave-type distribution, even for a single electron, but it can be observed more easily if many electron collisions have been recorded.

Calculating the wavelength of a particle

If an electron travels at a velocity of 1.000 × 10 7 m s –1 and has a mass of 9.109 × 10 –28 g, what is its wavelength?

Solution

We can use de Broglie’s equation to solve this problem, but we first must do a unit conversion of Planck’s constant. You learned earlier that 1 J = 1 kg m 2 /s 2 . Thus, we can write h = 6.626 × 10 –34 J s as 6.626 × 10 –34 kg m 2 /s.

λ = h m v
= 6.626 × 10 −34 kg m 2 /s ( 9.109 × 10 −31 kg ) ( 1.000 × 10 7 m/s ) = 7.274 × 10 −11 m

This is a small value, but it is significantly larger than the size of an electron in the classical (particle) view. This size is the same order of magnitude as the size of an atom. This means that electron wavelike behavior is going to be noticeable in an atom.

Check your learning

Calculate the wavelength of a softball with a mass of 100 g traveling at a velocity of 35 m s –1 , assuming that it can be modeled as a single particle.

Answer:

1.9 × 10 –34 m.

We never think of a thrown softball having a wavelength, since this wavelength is so small it is impossible for our senses or any known instrument to detect (strictly speaking, the wavelength of a real baseball would correspond to the wavelengths of its constituent atoms and molecules, which, while much larger than this value, would still be microscopically tiny). The de Broglie wavelength is only appreciable for matter that has a very small mass and/or a very high velocity.

Got questions? Get instant answers now!

Werner Heisenberg considered the limits of how accurately we can measure properties of an electron or other microscopic particles. He determined that there is a fundamental limit to how accurately one can measure both a particle’s position and its momentum simultaneously. The more accurately we measure the momentum of a particle, the less accurately we can determine its position at that time, and vice versa. This is summed up in what we now call the Heisenberg uncertainty principle    : It is fundamentally impossible to determine simultaneously and exactly both the momentum and the position of a particle . For a particle of mass m moving with velocity v x in the x direction (or equivalently with momentum p x ), the product of the uncertainty in the position, Δ x , and the uncertainty in the momentum, Δ p x , must be greater than or equal to 2 (recall that = h 2 π , the value of Planck’s constant divided by 2 π ).

Questions & Answers

what is d meaning of organic chemistry
Elizabeth Reply
it's a compound that comprises of hydrocarbon
Omoru
what iz alkanol
icha Reply
alkanol there are organic compounds with the functional group of ROH and relative molecular formula (CnH2n+1+OH)
Omoru
ok tnk u
icha
you are welcome
Omoru
What's alkaline soil
Amoo Reply
a system in which only energy is transferred between the system and the surrounding is called?
Ani Reply
which Element exhibit diagonal relationship with aluminum
Ani
following processes: Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and chlorine gas b. Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper(II) sulfate
Hisham Reply
How to know periodic table oftend
Ahmed Reply
u can know it through singing it as song it simple
Elizabeth
how to get atomic number of an element
Ogunleye Reply
how do you solve the examples in a much more explanatory way
Ogunleye
it seems by multiplying d number of d element by 2
Elizabeth
E.g like carbon 6*2=12 so d atomic number is 12
Elizabeth
The reaction of aceto nitrile with propane in the presence of the acid
Explain this paragraph in short
Manish Reply
What is solid state?
Manish Reply
What is chemical reaction
Manish
transforming reactants to product(s)
Andre
process
Andre
solid state is composed of tightly particles and it has a definite shape and volume
Elizabeth
Example of Lewis acid
Chidera Reply
Example of Lewis acid
Chidera
Chlorine
Mikidad
Anything with an empty orbital... the hydrogen ion is the most common example. BH3 is the typical example, but any metal in a coordination complex can be considered a Lewis acid.
Eszter
okay thanks
Jovial
aluminium and sulphur react to give aluminium sulfide.How many grams of Al are required to produce 100g of aluminium sulphide
Soni Reply
aluminium and sulphur react to give aluminium sulphide how many grams of Al are required to produce 100g of aluminium sulphide?
Soni
aluminium and sulphur react to give aluminium sulphide how many grams of Al are required to produce 100g of aluminium sulphide?
Soni
2Al+3S=Al2S3
galina
m(Al)=100×27×2/150=36g
galina
150 comes from?
Soni
thank you very much
Soni
molar mass of Al2S3
galina
150.158
thiru
Why can't atom be created or destroyed
Jacaranda Reply
matter simply converts to pure energy
that's nice
Meshach
explain how to distinguish ethanol from a sample of ethanoic acid by chemical test
Alice Reply
explain how ethanol can be distinguished from ethanoic acid by chemical test
Alice
Using a suitable experiment, describe how diffusion occurs in gases.
Melody Reply
when the excited energy which are in gaseous state collides with another to liberate from one place to another
Meshach

Get the best Chemistry course in your pocket!





Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

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

Ask