What are the unknowns in this nuclear reaction sequence?

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AStaunton
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Hi there

problem is:

given the following nuclear reaction sequence, determine X and Y:

[tex]_{9}^{15}N+_{1}^{1}H\rightarrow_{8}^{16}O+\gamma[/tex]

[tex]_{8}^{16}O+_{1}^{1}H\rightarrow_{9}^{17}F+X[/tex]

[tex]_{9}^{17}F\rightarrow_{8}^{17}O+Y+\nu_{e}[/tex]

My Attempted answer is:

As far as I can see, Y must be a [tex]e^{+}[/tex] in order to balance with the [tex]\nu_{e}[/tex] in the equation. But I am in doubt about this as the above way I am think is beta+ emission, however I read that the way of decay of fluorine-17 is electron capture so I don't know how to reconcile this.

Regarding what X is, I am not sure...I do not think it can be a [tex]\gamma[/tex] as binding energy per nucleon of oxygen 16 is greater than that of fluorine 17..I don't see how it can be a lepton as the lepton number seems to be balanced already...

Any feedback on my queries for X and Y is appreciated
 
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AStaunton said:
Hi there

problem is:

given the following nuclear reaction sequence, determine X and Y:

[tex]_{9}^{15}N+_{1}^{1}H\rightarrow_{8}^{16}O+\gamma[/tex]

[tex]_{8}^{16}O+_{1}^{1}H\rightarrow_{9}^{17}F+X[/tex]

[tex]_{9}^{17}F\rightarrow_{8}^{17}O+Y+\nu_{e}[/tex]

My Attempted answer is:

As far as I can see, Y must be a [tex]e^{+}[/tex] in order to balance with the [tex]\nu_{e}[/tex] in the equation. But I am in doubt about this as the above way I am think is beta+ emission, however I read that the way of decay of fluorine-17 is electron capture so I don't know how to reconcile this.

Regarding what X is, I am not sure...I do not think it can be a [tex]\gamma[/tex] as binding energy per nucleon of oxygen 16 is greater than that of fluorine 17..I don't see how it can be a lepton as the lepton number seems to be balanced already...

Any feedback on my queries for X and Y is appreciated

Hi !

I agree with Y must be a e^(+) (positron).

Because with Soddy's Law you have: For protons: 9 = 8 + Y + 0 <=> Y = 1
And for nucleons you have: 17 = 17 + Y + 0 <=> Y = 0
It's a positron !

Try to use my method.

Good luck ! :D
 
AStaunton said:
Regarding what X is, I am not sure...I do not think it can be a [tex]\gamma[/tex] as binding energy per nucleon of oxygen 16 is greater than that of fluorine 17..I don't see how it can be a lepton as the lepton number seems to be balanced already...
Extra energy can be supplied by kinetic energy, since the H impinges on the O.