Numerical Estimation of the momentum of free electrons

In summary: Yes. That's exactly what estimation is. How accurate do you think you need to be with this?"I think I need to be reasonably accurate with my estimate.
  • #1
hasan_researc
170
0

Homework Statement



This is a problem in numerical estimation.
Estimate a typical maximum momentum pmax for the free electrons in a metal where
pmax = [tex] \left(\frac{3nh^{3}}{8\pi}\right)^{1/3}[/tex]

(In estimating n, the number density of free electrons, assume there is one such electron associated with every atom in the metal, and either look up or guestimate how dense
typical metals are compared with water – for which you should know the density.)

Homework Equations



Density of water = 1000 kg/m3
Molar mass of water = 18 g/mol
Molar mass of iron = 55 g/mol (Iron, I suppose, exhibits the typical properties of metals.)

The Attempt at a Solution



Molar mass of iron/Molar mass of water = 3. Therefore, one iron atom is 3 times heavier than a water molecule.

But I do not know anything about the packing of atoms/molecules in iron or water.
SO, how do I guesstimate the number density of free electrons in a free metal?

Anyway, if I proceed in my calculation with n = 1028, then

pmax = [tex]\left(\frac{3nh^{3}}{8\pi}\right)^{1/3}[/tex]
= [tex]\left(\frac{3*10^{28}*(6.63*10^{-34})^{3}}{8\pi}\right)^{1/3}[/tex]
= [tex](6.63*10^{-34})\left(\frac{3*10^{28}}{8*3}\right)^{1/3}[/tex]
= [tex] (6.63*10^{-34})\left(\frac{10^{28}}{8}\right)^{1/3}[/tex]
= [tex]\left(\frac{(6.63*10^{-34})*10^{9}}{2}\right)[/tex]
= [tex]3.3*10^{-25}[/tex]

Does the answer look reasonable. How can the estimation working be improved?

Any help would be greatly appreciated.
 
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  • #2
You could easily look up the actual density (mass/volume) of iron and calculate the electron density from that...
 
  • #3
I surely can, but I want to find the density using my own reasoning (w/o help from external sources).
 
  • #4
The problem statement says to either look up or guesstimate the density of a typical metal. Can you estimate a metal's density, relative to water, from your own personal experience? I.e., is it heavier or lighter than water? By what factor, roughly?

If you can't make an estimate you are reasonably confident about, my advice is to look it up.
 
  • #5
Thank you for your help!

" Can you estimate a metal's density, relative to water, from your own personal experience?" :

I'm not just sure whether using personal experience is a good method to estimate the density of a typical metal !? Otherwise, I could easily quote a factor (eg, 3 or 4).

What I want to be able to do is work out the metal density relative to water density by comparing the atomic structure and properties of metals and water. Unfortunately, my knowledge is not adequate to answer that. So, I am hoping someone could help me with that.

[I can easily look up the value in a book, but then I will not have learned how to guesstimate ths quantity, so...]
 
  • #6
hasan_researc said:
I'm not just sure whether using personal experience is a good method to estimate the density of a typical metal !? Otherwise, I could easily quote a factor (eg, 3 or 4).
Yes. That's exactly what estimation is. How accurate do you think you need to be with this?
hasan_researc said:
What I want to be able to do is work out the metal density relative to water density by comparing the atomic structure and properties of metals and water.
This would be a detailed calculation, not an estimation. If you want to know how to do it, I'd suggest you head over to the General Physics forum and ask how you can calculate the density of iron from its crystal structure, or the density of water from its molecular bonding properties. But I can't imagine that anyone would actually calculate the density of iron this way if all they need is a numerical value for that density.
 

FAQ: Numerical Estimation of the momentum of free electrons

What is numerical estimation of the momentum of free electrons?

Numerical estimation of the momentum of free electrons is a method used to calculate the momentum of electrons that are not bound to an atom or molecule. It is an important concept in quantum mechanics and is often used in various fields such as physics, chemistry, and engineering.

Why is it important to estimate the momentum of free electrons?

Estimating the momentum of free electrons is important because it allows us to understand the behavior of electrons in different environments and how they contribute to various physical and chemical processes. It also helps us to design and engineer devices that utilize the properties of electrons, such as transistors and microchips.

How is the momentum of free electrons calculated?

The momentum of free electrons can be calculated using the formula p = mv, where p is the momentum, m is the mass of the electron, and v is the velocity. In quantum mechanics, the momentum of an electron can also be represented by its wavefunction, which is a mathematical expression that describes its wave-like behavior.

What factors affect the momentum of free electrons?

The momentum of free electrons can be affected by several factors, including the speed and direction of the electron, the strength of the electric and magnetic fields it is exposed to, and the presence of other particles or objects that interact with the electron. Additionally, the mass of the electron itself also plays a role in determining its momentum.

How is numerical estimation used in the study of free electron momentum?

Numerical estimation is often used in conjunction with experimental data to validate and refine theories and models related to the momentum of free electrons. It also allows us to make predictions and calculations for systems that are too complex to be solved analytically. Additionally, numerical methods are used to simulate and visualize the behavior of electrons in different scenarios, providing valuable insights into their properties and interactions.

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