Why is there a difference ?Melting Point of Lead & Tungsten: Find RMSA

In summary, the conversation is about finding the root mean square amplitude of atomic vibrations for lead and tungsten, using Lindemann's criterion as an approximation. The approach involves finding the vibrational energy and equating it to 1/2kx^2, and using equations to calculate the spring constant and separation at equilibrium. There is some confusion about the variation of Lindemann's constant and the exact definition of melting. The conversation also mentions different sources with conflicting equations for mean square amplitude and the expression for interatomic distance.
  • #1
ibysaiyan
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Homework Statement


Hi all, I have a question which is regarding melting point. I know that lindemann's criterion doesn't actually describe the phenomena of melting but is an approximation of what maybe occurring/ an intuitive approach.

So the question is in a tabulated form for lead and tungsten ( assuming crystals) with Youngs modulus, Molar volume and melting temperature given and they want me to describe the terms / details quantitatively.

I have assumed that they want me to essentially find out the root mean square amplitude of the vibrations which will eventually give me a separation distance (not equlibirum) that needs to be at least 10% of a fraction to the inter atomic distance.

Homework Equations

Root mean square of amplitude of atomic vibrations is given by the following equation:

[itex]\sqrt{<u^2>}[/itex] = [itex]\sqrt{KbT/D}[/itex] where D = spring constant.

D = Y[itex]r_{0}[/itex]

The Attempt at a Solution



Data given for lead is:

[itex]T_{m}[/itex] = 3650 K , Molar volume : 0.0953m^3 , Young's modulus: 36*10^10 N/m^2

My approach involved finding out the vibrational energy by kbT. Then equating that to 1/2kx^2 ( here 'x' = r-r0) but first I need to find out spring constant k and separation at equlibirum.

To get 'k' I have used the following equation:
k= Y[itex]r_{0}[/itex]
and for [itex]r_{0}[/itex] , Molar volume = [itex]N_{a}[/itex]b* ([itex]r_{0}[/itex])^3Is the above approach right ? is this how I am mean't to do it ?

Thanks for your replies and merry christmas !Edit: Also one more thing which confuses me slightly .. I know that lindemann's constant is a value which's usually same i.e doesn't vary but I have seen different variation to get it... is the rms of the amplitude of the vibration = C * interatomic distance ? can someone clarify this for me as well. Thanks!

EDIT2: If you have a look on the following links.. you will notice that there is a difference of powers for the mean square amplitude value <u^2> = ..
http://phycomp.technion.ac.il/~phsorkin/thesis/node4.html
http://en.wikipedia.org/wiki/Melting_point
 
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  • #2
Can someone clairfy to me when exactly does melting occur. Is it when the root mean square of the amplitude gets to about 10% of the inter atomic distance or is it just the mean square of the amplitude ?

Also to find the inter atomic distance.. is that same as 'r0' separation of atoms at equilibrium ?

<u^2> = Cl^2 * a^2 is this the right expression ? Google hasn't helped much on this but has shown some conflicting form.
Thank you..
 

Related to Why is there a difference ?Melting Point of Lead & Tungsten: Find RMSA

1. Why is the melting point of lead lower than that of tungsten?

The melting point of a substance is determined by the strength of the bonds between its atoms. Lead has a lower melting point because it has weaker metallic bonds compared to tungsten. This means that it takes less energy to break these bonds and melt the lead.

2. What is the significance of the difference in melting points between lead and tungsten?

The difference in melting points between lead and tungsten is significant because it reflects the difference in their physical properties and behaviors. Lead, with its lower melting point, is more malleable and easier to work with, making it useful for various applications such as batteries and piping. Tungsten, on the other hand, with its high melting point, is extremely hard and has a high resistance to heat, making it suitable for use in high-temperature environments like light bulbs and rocket engines.

3. What is the relationship between the melting point of a substance and its density?

There is a general trend that substances with higher melting points also have higher densities. This is because the stronger bonds between the atoms require them to be packed more closely together, resulting in a higher density. This is true for both lead and tungsten, with tungsten having a higher density due to its higher melting point and stronger metallic bonds.

4. How does the melting point of lead and tungsten relate to their atomic structures?

The atomic structure of a substance plays a significant role in determining its melting point. In the case of lead and tungsten, both are metallic elements with similar crystal structures, known as body-centered cubic (BCC). However, tungsten has a higher atomic number and a larger atomic radius, which leads to stronger metallic bonds and a higher melting point compared to lead.

5. How does the RMSA (Root Mean Square Atomic) of a substance affect its melting point?

The RMSA of a substance refers to the average distance between its atoms. It is a measure of the packing efficiency of the atoms in a crystal structure. In general, substances with a higher RMSA have weaker bonds and thus lower melting points. This is true for lead and tungsten, as lead has a larger RMSA and weaker bonds, resulting in a lower melting point compared to tungsten.

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