What are the unknowns in this nuclear reaction sequence?

AI Thread Summary
The discussion centers on determining the unknowns X and Y in a nuclear reaction sequence involving nitrogen, hydrogen, oxygen, and fluorine isotopes. Participants agree that Y is likely a positron (e+) to balance the equation, referencing Soddy's Law for protons and nucleons. There is uncertainty regarding X, with suggestions that it cannot be a gamma photon due to binding energy considerations. The conversation highlights the complexities of decay processes, particularly the electron capture of fluorine-17, which adds to the confusion about the reactions. Overall, the participants seek clarity on these nuclear reactions and the implications of energy balance.
AStaunton
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Hi there

problem is:

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

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

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

_{9}^{17}F\rightarrow_{8}^{17}O+Y+\nu_{e}

My Attempted answer is:

As far as I can see, Y must be a e^{+} in order to balance with the \nu_{e} 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 \gamma 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:

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

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

_{9}^{17}F\rightarrow_{8}^{17}O+Y+\nu_{e}

My Attempted answer is:

As far as I can see, Y must be a e^{+} in order to balance with the \nu_{e} 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 \gamma 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 \gamma 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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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