1
$\begingroup$

$$\ce{_2^4He + _13^27Al -> _15^30P + _0^1n}$$

Will this reaction qualify as a nuclear fusion reaction? The answer to this question in the test says that it won't, yet it seems to be the reverse of a fission reaction. Why would it not be considered fusion?

| improve this question | |
$\endgroup$
  • 2
    $\begingroup$ While it is certainly a nuclear reaction, but nuclear physics folks would likely just call it an ($\alpha$,n) reaction. $\endgroup$ – Jon Custer Nov 17 '17 at 16:41
  • 2
    $\begingroup$ Were you given a definition of nuclear fusion reaction in class? As far as I can tell, this would be an example of fusion. $\endgroup$ – Tyberius Nov 17 '17 at 17:58
  • $\begingroup$ Either the test answer is wrong, or the answer depends on some weird(?) definition of a fusion reaction. It sure seems like a fusion reaction to me. $\endgroup$ – MaxW Nov 17 '17 at 18:38
  • 1
    $\begingroup$ Well, there are two elementary steps here - one is fusion, later neutron emission happens. $\endgroup$ – Mithoron Nov 17 '17 at 23:58
  • $\begingroup$ It may be that your teacher's definition of a fusion reaction is one that releases energy. This reaction requires significant energy input. Admittedly, that isn't a good definition of a fusion reaction. $\endgroup$ – matt_black Nov 18 '17 at 11:00
8
$\begingroup$

The Wikipedia article on Nuclear Fusion starts off:

In nuclear physics, nuclear fusion is a reaction in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles (neutrons or protons). The difference in mass between the reactants and products is manifested as the release of large amounts of energy. ...

$\ce{^4He}$ and $\ce{^{27}Al}$ are certainly different than $\ce{^{30}P}$.

The masses of the products however are more than the masses of the reactants

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2} \begin{array}{lrl} \text{4-He} & 4.&\hspace{-1em}00260325415 \\ \text{27-Al} & 26.&\hspace{-1em}98153863\\ \hline \text{total} & 30.&\hspace{-1em}98414188\\ \end{array}$$

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2} \begin{array}{lrl} \text{30-P} & 29.&\hspace{-1em}9783138 \\ \text{n} & 1.&\hspace{-1em}00866491588 \\ \hline \text{total} & 30.&\hspace{-1em}9869787 \\ \end{array}$$

So it would seem that the definition used would require the products to have less mass than the reactants. This seems like a weird definition to me...

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Not really, fusion can make nuclei with less binding energy, but this energy has to be provided, much like in chemistry. See my comment^ $\endgroup$ – Mithoron Nov 18 '17 at 0:01
  • $\begingroup$ @Mithoron - I agree with you. It seems like a fusion reaction to me. However in order for the reaction that the OP gave not to be a fusion reaction, per test answer, the mass deficit is the only reason that I can think of. It depends of course on how the particular professor defines a fusion reaction. $\endgroup$ – MaxW Nov 18 '17 at 0:54
  • $\begingroup$ Okay, so between two such reactions, the one that mentions a release of energy is more appropriately a fusion reaction, correct? $\endgroup$ – Shreya Nov 19 '17 at 5:51
  • 1
    $\begingroup$ @Shreya - I would consider the reaction given a type of fusion reaction. For various types see en.wikipedia.org/wiki/… I was trying to explain why the test answer was marked that it wasn't a fusion reaction. So you need to clarify the exact definition your instructor used for a "fusion reaction." $\endgroup$ – MaxW Nov 19 '17 at 5:57

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.