# Homework Help: Simple Q-value calculation for n+14N -> 12C + t

1. Dec 23, 2012

### Silversonic

1. The problem statement, all variables and given/known data

Apologies if this is in the wrong section, it seemed too basic to place in the advanced physics bit. I'm just trying to calculate the Q-value for the reaction process

n + 14N → 12C + 3H

3. The attempt at a solution

The Q-value is just the difference in the nuclear masses on each side of the equation, I do some re-arranging to find out it's actually the difference in the mass excess of each atom on both sides of the equation. I'm using this site for the mass excesses;

http://www.nndc.bnl.gov/amdc/masstables/Ame2003/mass.mas03 [Broken]

For the neutron, 14N, 12C and 3H (tritium) the mass excesses are 8071.31710, 25251.164, 24926.178 and 14949.80600 keV respectively. I plug this in, and get the Q-value as -6.55 MeV, but I know the answer is meant to be -4.015MeV. Indeed, I've been told it is and also this Q-value calculator;

Confirms it for me. So what could I be doing wrong? I'm definitely not typing it in to my calculator incorrectly!

Edit: Aside note, I'm told that the neutron must have the threshold energy of 4.015MeV in the CM frame for this process to occur. Is there any reason it's specific to the CM frame? Is it to do with the fact that the total energy of the system in the CM frame is the minimal energy as seen from all possible inertial frames?

Last edited by a moderator: May 6, 2017
2. Dec 24, 2012

### Staff: Mentor

Did you check the calculation with the masses of the nuclei? Maybe one of the values is wrong.

Neutron energy is frame-dependent, so you need some specific frame to give the energy. You could use the frame of the target as well.

3. Dec 24, 2012

### BruceW

I agree with your mass excesses for the neutron and tritium (assuming that they are stationary). But I get something different for the mass excesses of 14N and 12C (Assuming that they too are stationary). The mass excess for 12C is simple, if you think about it. And I get something close to your mass excess for 14N, but not quite the same.

Are you sure it is meant to be 4.015MeV in the centre of momentum frame? Don't you mean in the target frame? Think about how you are calculating your Q-value, you are assuming that none of the objects are moving, right? Then you are told that the Q-value will be -4.015MeV So for the energy to balance, you would need an extra 4.015MeV on the left hand side. Where could you get this energy from?

4. Dec 24, 2012

### Silversonic

I see my mistake. It's 14N and 12C, where 14 and 12 refer to the mass numbers. I was looking up Nitrogen and Carbon with 14 and 12 neutrons respectively (with mass numbers of 21 and 18 respectively). I was doing this late at night. I wonder sometimes how I put my clothes on in the morning....all by myself...it's amazing!

But yeah, it's clear that the mass excess of 12C should be 0 since the atomic mass unit is measured relative to carbon-12, it's a twelfth of its mass. Should've realised that at the time.

Hmm, I look at it this way. The Q-value is the energy released from the binding energy due to the nuclear reaction. This energy is shared into the KE of the constituents. Thus, if the projectile/target have no KE, the Q-value is directly equal to the sum of the KE of the produced particles. Hence, in this case, it can't make any sense for their sum to be negative.

When the projectile has KE, the Q-value is equal to the sum of the KE of the produced minus the KE of the projectile. Hence, if the KE of the neutron (projectile) is less than 4.015MeV, the sum of the KE of the produced must still be negative (still not making sense). At the very least for this reaction to occur and the KE sum of the produced to be positive or zero, the KE of the neutron must be equal to the Q-value.

However, to answer mfb/Bruce about the target frame. Doesn't the energy of a system depend on what inertial frame you're measuring it in? What if, for example, the target and the projectile were moving in the same direction with NEARLY the same speed (the neutron being faster) such that the sum of their KEs in the stationary lab frame is enough to allow the reaction process to be possible. In the Nitrogen's frame of reference the neutron is traveling very slowly, and doesn't have enough energy to produce the reaction.

Does this mean that for a reaction to occur, the reaction must be energetically possible in ALL frames of reference? And that if it's possible in one frame of reference it doesn't necessarily mean it occurs, since it may not be possible in another frame of reference?

Wikipedia reveals to me that the CoM frame is the frame in which the energy of the system is the least, hence if it's possible in that frame it's possible in all other frames. Being possible in the target frame doesn't necessarily mean it's possible in the CM frame?

Last edited: Dec 24, 2012
5. Dec 24, 2012

### Staff: Mentor

Right, and you get the lowest energy in the center of mass frame. This is useful for calculations, as you can assume that the products are at rest afterwards at threshold - you do not have to consider their kinetic energy. In all other frames, the products cannot be at rest (momentum conservation) and you have to calculate their kinetic energy.
The sum of kinetic energy of the neutron and N14 should be 4.015MeV - in the center of mass system, most of this energy comes from the neutron.

If it is possible in any frame, it is possible in all. "Reaction is possible" is coordinate independent.