Calculating Kinetic Energy of Reaction Products in 13-C (d,n) 14-N Reaction

AI Thread Summary
The discussion focuses on calculating the total kinetic energy (KE) of the products from the 13-C (d,n) 14-N reaction, given an incoming deuteron with KE of 36.3 MeV. Participants emphasize using the conservation of energy principle, suggesting that the total energy before the reaction equals the total energy after, incorporating the Q-value concept. There is confusion regarding the application of E=mc^2 and the need for velocity to determine individual particle kinetic energies. It is clarified that the initial KE of the deuteron is indeed 36.3 MeV, and participants suggest using classical mechanics to find velocity. The discussion highlights the complexity of the problem, indicating it may be challenging for a high school physics level.
f4d_girl
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"Calculate the total KE of the products of the reaction

13-C (d,n) 14-N if the incoming deuteron (d) has KE = 36.3 MeV"

I used the Q-value equation which is

Q = KEn + KEN - KEn - KEd

but to find the kinetic energy for each particle, velocity is required (which isn't given in the question)

So, i tried to use E=MC^2 and didn't work out well

to calculate the total KE of the products (KE of neutron and nitrogen)

can't i just use E=MC^2?

please help:confused:

Thanks
 
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One has to use conservation of total energy, i.e the sum of kinetic and rest energies before = sum of kinetic and rest energies after.

The Q value is the difference in rest energy (masses). If Q > 0, the released energy is manifest in the kinetic energy of the products.

If Q < 0, some energy (e.g. kinetic energy of one or both reactants) must be applied.
 
umm okay
so you're saying that i should use KE before = KE after?
but the question is how do i find a kinetic energy of Carbon, neutron and nitrogen?
using E=mc^2?
 
Try KE (after) = KE (before) + Q or

KE(final) = KE(initial) + Q

and one can use classical mechanics for kinetic energy since 36.3 MeV << 1875.6 MeV (rest mass of d).

If one wants to calculate the specific kinetic energy of the particles, then one must apply conservation of momentum (a vector quantity) in both the x and y or longitudinal and transverse directions taken with respect to the incident velocity of the deuteron. A reasonable assumption would be that the beam of deuterons is impinging upon a fixed (solid) target of C.
 
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what I'm asking is, is KE(before) = 36.3 MeV?

KE= 0.5 mv^2, but velocity isn't given in the question
 
i still didn't get it
perhaps this is too complicated question for physics grade 12 IB
but thanks anyway
 
f4d_girl said:
what I'm asking is, is KE(before) = 36.3 MeV?

KE= 0.5 mv^2, but velocity isn't given in the question
You are given the kinetic energy of 36.3 MeV, which is 1/2 mv2.

You could calculate v = sqrt(2KE/m).

Try this page and browse the site - http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html
 
i've never heard of this equation v = sqrt(2KE/m)
but i'll try!
thanks
 
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