Binding energy Help please

[Bolter]

Homework Statement:

Working out how to calculate binding energy

Relevant Equations:

l = 1/2Nne

Here is the question:

Stuck on how to complete part c)ii

Here is what I have done so far as well as trying to answer part c)ii

Any help would be appreciated! Thanks

 

[kuruman]

37 kJ = 37x10³ J.

 

[Bolter]

37 kJ = 37x10³ J.

Oh shoot yes you’re right. I don’t know why I processed it at as mega joules

So doing the following calculation again gives

Binding energy = (2 * 293000) / (6x10^23 * 7) = 1.41x10^-19 J which is closer to the answer but still not right

 

[mjc123]

296 kJ/kg. You need kJ/mol.

 

[Bolter]

296 kJ/kg. You need kJ/mol.

So to convert from kJ/kg to kJ/mol, I must multiply kJ/kg by the molar mass?

I did this and got the following

Binding energy = 2(293000 * 0.20)/(6x10^23)(7) = 2.79... x10^-20 J which is roughly equal to 2.8x10^-20 J

I also got another part that I need help with if you don’t mind

Not entirely sure how to approach part d)ii but I suppose I had done part d) i right

 

[mjc123]

di is right. Not sure what the problem is for dii - find the volume per atom, as you did for Hg.

 

[Bolter]

di is right. Not sure what the problem is for dii - find the volume per atom, as you did for Hg.

To find volume per atom. I know that 1.0691... x10^-4 moles of Hg is contained in 2x10^-4 m^3.

So to firstly get the volume per mole of Hg, I do ' 2x10^-4 / 1.0691...x10^-4 ' which gives 1.870... volume per mole

Dividing 1.870... by avogrado's constant is '1.870... / 6.022x10^23 = 3.106... x 10^-24 volume per atom
Cube rooting volume per atom number should give the spacing which comes out to be '1.459... x10^-8 m'

Would you agree that these steps are the right way of doing this?

 

[mjc123]

Yes, except that you're talking about neon, not mercury.

 

[Bolter]

Yes, except that you're talking about neon, not mercury.

Yes, sorry I got mixed up with the wording somehow

 

[mjc123]

The other thing is that I would be more explicit about tracking units through the calculation. So rather than say "1.870 volume per mole" I would say "1.870 m³ volume per mole"; likewise "3.106 x 10^-24 m³ volume per atom". In this case, as you're dealing in m all the time, it's not so critical, but if you're switching between e.g. m and cm, or J and kJ, you can easily make mistakes if you don't keep track of units.