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I wanted to know whether the bombarding a particle to transmute it into another particle releases or absorbs energy. For eg, I wanted to know specifically for O-17 formation:

O-17 formation from the bombardment of Nitrogen-14 with alpha particles

I found a question which said that the mass defect of this reaction is negative so energy is in fact taken in. But O-17 is heavier than N-14 and my own calculations show that the mass defect of Nitrogen is 0.1039 AMU and mass defect of Oxygen-17 is 0.1358 AMU. In this case, I have studied that since the resultant nucleus has a higher mass defect, the binding energy is higher -- which is released during formation and there is a net energy release. Is this true or not? Or is this reaction endothermic?

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mass defect of this reaction is negative

This is a double negative statement (defect and negative) which means that the mass of the products on the right hand side of the equation is greater than the mass of the reactants on the left hab=nd side of the equation .

If you use a table of isotopic masses you can find out what this mass is by subtracting the combined masses of the reactants from the combined masses of the products.

If this reaction is to occur the reactants must have a least the energy equivalent of the negative mass defect, using $\Delta E = mc^2$, as kinetic energy so in chemistry parlance this is an endothermic reaction.

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  • $\begingroup$ Your general answer is correct but I'd just like to point out that if you indeed use the table for atomic masses you will not get the correct answer of having more mass on the right hand side. Hence you need a table of nuclear masses. I know this because I've used that exact link and failed to come to the proper conclusion but after some long searching to find nuclear masses I could reach the proper answer. $\endgroup$ – Tausif Hossain Aug 2 '18 at 9:12
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    $\begingroup$ @TausifHossain The table of atomic masses will yield the correct answer as the number of electrons on the lhs (9) is the same as the number of electrons on the rhs (9) and the binding energy of the orbitting electrons is much, much less than the binding energy of the nucleons in the nucleus. Using the table that I cited the increase in mass is 0.00128 u. $\endgroup$ – Farcher Aug 2 '18 at 9:30
  • $\begingroup$ I understand now, the discrepancy is because I took the atomic mass of N-14 instead of nuclear mass, so correcting it. $\endgroup$ – Tausif Hossain Aug 2 '18 at 9:33
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    $\begingroup$ www-nds.iaea.org/amdc is a good source, including a Q-value calculator. $\endgroup$ – Jon Custer Aug 2 '18 at 15:21
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The reaction

14N + 4He = 17O + 1H,

let us calculate the Q-value of the reaction.

The amount of energy per nuclear reaction.

Masses(in amu): H = 1.007825; n = 1.008665;

He= 4.00260; 14N = 14.00307; and 17O = 16.99914.

Conversion factor : 1 amu = 1.66053.10^-27 kg, c = 3.0.10^8 m s^-1 (velocity of light).

The reaction 14N (He, p) 17O is endothermic as supported by its

Q = -1.207 MeV/reaction (negative indicating energy required)

ref.http://www.science.uwaterloo.ca/~cchieh/cact/nuctek/assgn8a.html

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