What Is the Nuclear Binding Energy of N-14 in Joules per Mole?

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Discussion Overview

The discussion revolves around the calculation of the nuclear binding energy of nitrogen-14 (N-14) in joules per mole. Participants are exploring the conversion of mass deficit into energy using Einstein's equation, while addressing issues related to unit conversions and the mass of nitrogen atoms.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant inquires about the nuclear binding energy of N-14 and describes their approach to calculating it using the mass deficit and Einstein's equation, but expresses uncertainty about their method.
  • Another participant questions the conversion of mass deficit into kilograms and prompts clarification on the mass of a mole of nitrogen atoms.
  • A subsequent reply suggests that one mole of nitrogen is 28 grams, but another participant corrects this by stating that one mole of nitrogen atoms is actually 14 grams, specifically 14.00307 grams, emphasizing the distinction between atomic weight and mass in grams versus kilograms.
  • There is a challenge regarding the correct interpretation of the mass deficit in relation to the units used in the calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct conversion of mass deficit or the appropriate units to use in the calculations. Multiple views on the mass of nitrogen and its implications for the binding energy calculation remain unresolved.

Contextual Notes

There are limitations regarding the assumptions made about unit conversions and the definitions of mass in grams versus kilograms, which are not fully resolved in the discussion.

bree1622
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What is the nuclear binding energy of N-14 in joules per mole of nuclei?
The mass deficit is 0.11278. Atomic mass is 14.00307. Total mass is 14.11585.
I keep getting stuck after converting and substituting it into the equation.
First I converted the mass deficit into kilograms. 0.11278 =1.1278 × 10^-1KG. Then I put into equation. E=mc2=1.1278 × 10^-1 kgmol × (2.998×108ms)^2 . I am not quite sure if this is right so far or what to do next. Any help is greatly appreciated. Thanks!
 
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bree1622 said:
First I converted the mass deficit into kilograms. 0.11278 =1.1278 × 10^-1KG.
Is that so? What is the mass of a mole of nitrogen atoms, for example?
 
mjc123 said:
Is that so? What is the mass of a mole of nitrogen atoms, for example?
Is one mole of nitrogen 28g?
 
I said one mole of nitrogen atoms, so the answer is 14 g, or to be more precise 14.00307 g. But the point is the same. The atomic weight is the mass in g (not kg) of 1 mole. So the mass deficit is not 1.1278 x 10-1 kg/mol.
 

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