Electronvolt to joule/mole

  • #1
5
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Main Question or Discussion Point

Dear Friends,

Could you tell me how I can convert electronvolt to joule/mole for silicon material? I want to convert 4530 ev to joule/mole for silicon.

Many thanks in advance.
 

Answers and Replies

  • #2
31
0
The trick I like to use is to "substitute" the units.

Recall that 1.6*10^-19 J = 1 eV

We can write 6.7 eV as 6.7 * 1eV. Substitute the 1eV from earlier and you get 6.7*1.6*10^-19 J.

I'll leave the rest as an exercise, but if you get stuck, try thinking for how much silicon the 4530eV is for.
 
  • #3
5
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Thank you for your respone.

I calculated it. Is that correct?

number of atoms = 27000
silicon mass = 28.0855*1.66054*10^-27 Kg
1 Kg silicon = 35.6 mole

(4530*1.6*10^-19)/(27000*28.0855*1.66054*10^-27 ) = 600 KJ/Kg

600/35.6 = 16.85 KJ/mole
 
  • #4
31
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I have a few questions:

1) What is the energy of 4530eV for? Is it for some atomic transition?

2) Where did the number of 27000 atoms come from? Is the 4530eV for 27000 silicon atoms?
 
  • #5
5
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thanks for your help.

1) That is enthalpy when the solid melts.
2) 4530 ev is for 27000 atoms.
 
  • #6
31
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Yes, your calculation seems correct, but a few things to consider:

1) You might want to check the math again. I obtained 16.168 kJ/mol. The discrepancy might be because you approximated the value to be 600J/kg.

2) From wikipedia, we have:
1 kJ·mol−1 is equal to 0.239 kcal·mol−1 or 1.04×10−2eV per particle.
So, if this is one way to convert if you have good memory!

3) While your calculation is correct, I find it rather long. In fact, if you notice, 1/(35.6(mol/kg)*28.0855*1.66054*10^-27(kg))=6.023078*10^23 /mol, which is the Avogadro constant!

4) In view of point 3, in fact, you can directly compute using (check units by cancelling them like in fractions):
4530eV/270000atoms * 1.6*10^-19 (J/eV)* 6.022*10^23 (atoms/mol) = 16.168kJ/mol.
Indeed, 1.6*10^-19*6.022*10^23 = 96352 J/mol, which converts 1.04*10^-2eV to 1kJ/ mol in point 1!
 

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