Minimum Kinetic energy and Reaction Q-value

July 10, 2023

GoLuckyRyan Basic Leave a comment

just a note to myself.

The Q-value for a reaction A(a,b)B is defined as

\displaystyle Q = m_A + m_a - m_b - m_B

In general, the kinetic energy of the incoming beam must compensate the Q-value for the reaction to happen, i.e.:

\displaystyle m_A + T + m_a \geq m_b + m_B \Rightarrow T \geq -Q

This gives the idea that the beam energy should be larger than the negative of the Q-value. However, this is not quite true. This is because the kinematics energy is not minimum unless in the CM frame.

To prove this, using the non-relativistic transform, the kinematic energy of the system after transforming with velocity v is

\displaystyle T' = \frac{1}{2} m_A ( v_A - v)^2 + \frac{1}{2} m_a v^2

The minimum of T' is reached when v = (m_A)/ (m_a + m_A) v_A , which is the velocity to transform to the CM frame. The same result can be found for the relativistic case by minimizing the total energy with an arbitrary Lorentz boost

\displaystyle E' = \gamma E_A - \gamma \beta k_A + \gamma m_a

differential the above equation and find the minimum, we get \beta = k_A/(E_A + m_a) , which is the Lorentz boost to the CM frame.

Therefore, we should look at the energy balance in the CM frame.

The CM frame energy balance in non-relativistic is

\displaystyle m_A + m_a + \frac{m_a}{m_A+ m_a} T_A \geq m_b + m_B

where T_A is the KE of A in the lab frame. This gives

\displaystyle T_A \geq -Q \left( 1 + \frac{m_A}{m_a} \right)

For relativism, let’s simply quote the result

\displaystyle T_A \geq \frac{1}{2m_a} \left( (m_b + m_B)^2 - (m_a + m_A)^2 \right)

substitute the Q-value,

\displaystyle T_A \geq \frac{Q^2}{2m_a} - Q\left( 1+ \frac{m_A}{m_a} \right)

In the relativistic result, the term Q^2/ 2/ m_a is usually small.

The minimum KE per nucleon should not matter for Invert and normal kinematics.

\displaystyle T_{min} / A \approx \frac{m_\mu}{2m_a m_A} \left( (m_b + m_B)^2 - (m_a + m_A)^2 \right)

The interchange of particles a and A does not matter. Subitute the Q-value,

\displaystyle T_{min} / A \approx - Q \left( \frac{a + A}{a A} \right) + \frac{m_\mu Q^2}{2 m_a m_A} \approx - Q \left( \frac{a + A}{a A} \right)

We can make one more approximation, without loss of generality, whatever mass is smaller and the difference between a and A is large,

\displaystyle T_{min} / A \approx - \frac{Q}{ \min(a, A)}

Q-value and z-position in HELIOS

July 9, 2023

GoLuckyRyan Basic , , Leave a comment

in HELIOS, I always have a feeling that the Q-value is related to the z-position. Look at the Q-value,

\displaystyle Q = m_A + m_a - m_b - m_B - E_x,

where E_x is the excitation energy of the heavy recoil B. And we know that the higher the excitation energy, the z-position will shift toward the positive z-position. But that is for a given reaction. Is there any general or universal relationship between the Q-value and the z-position?

Instead of solving the equations and finding out the relation, I use simulation, put various reactions, and beam nuclei (6,9Li, 10B, 12C, 16N, 16,20O, 19F, 25Mg, 30,38Si, 40,48Ca, 96Zr, 158Tb, 208Pb, 238U ) adjusting the E_x from 0 to 2 MeV to mimic the change of Q-value. And I fixed the z-position at \theta_{cm} = 10 \circ , the beam energy of 10 MeV/u, and the magnetic field was 2.5 T. Here is the result:

For the given reaction, say the two neutron transfer reactions (d,p) and (t,d) has a similar trend. And (d,3He) and (d,a) are also on the different sections of the same curve. In general, all curve, except the (d,t), has a similar slope, roughly -32 mm/MeV, the (d,p) and (t,d) is a bit steeper with a slope of -37 mm/MeV.

Also, in the 2-neutron-adding (t,p) reaction, the z-position (-800 mm) at Q =0 is roughly double of the 1-neutron-adding (d,p) or (t,d) reaction, which is ( ~500 mm) . And charge exchange reaction (t, 3He) is almost at z = 0. The proton-removal reactions (d, 3He) and (t, a) share a similar verticle offset of 400 mm. and neutron-removal and pn-removal reactions have z-position at ~ 700 mm. In general, adding reactions has negative z, and removal reaction has positive z for zero Q-value.

Last note, with the higher magnetic field, the curves will move toward z = 0, and the slope will also change, for example, the (d,p) reaction

We can see the effect of the B-field is stronger for a higher Q-value reaction for (d,p) reaction. In general, we can foresee there is a Q-value that is unaffected by the B-field, and any Q-value further away from this special Q-value will be affected by the B-field the more. In fact, since the “contraction” is around z=0, the special Q-value that is unaffected by the B-field is near the curve at z=0. For (d,p) and (t,d) reactions, this special Q-value is around -9 MeV. For (t, 3He) reaction, this special Q-value is ~ 0 MeV. For (t, a) reaction, it is around 11 MeV.

And If I change the beam energy but keep the B-field to 2.5 T for the (d,p) reaction:

Higher the beam higher, the higher the energy of the light recoil, and the z-positon move further away.

Q-value and Binding energy

February 16, 2011

GoLuckyRyan Basic, Working on , , Leave a comment

I already talked on binding energy.

And the Q-value is the mass different between nuclear reaction. In same case binding energy is same as the Q-value.

Q-value is :

Q= \sum{m_i} - \sum{m_f}

By energy conservation, it can be rewritten by kinetic energy.

Q= \sum{T_f} - \sum{T_i}

But using energy is bit troublesome due to the frame transform. Using mass is simple, since it is a Lorentzian invariance.

However, during scattering experiment. The kinetic energy is much more easy to measure. Thus the kinetic form is used more frequently in experiment.

I will give the expression of K.E. in lab frame later.

2p-2p decay of 8C and isospin-allowed 2p decay of the isobaric-analog state in 8B

February 3, 2011

GoLuckyRyan experimental, papers reading, Working on , , , , , , , , , , , , Leave a comment

DOI: 10.1103/PhysRevC.82.041304

this paper reports another 2 protons decay mode in 8C. They also discover an “enhancement” at small relative energy of 2 protons. They also reported that an isobaric analog state, 8C and 8B, have same 2-protons decay, which is not known before.

the 1st paragraph is a background and introduction. 2 protons decay is rare. lightest nucleus is 6Be and heaviest and well-studied is 45Fe. the decay time constant can be vary over 18 orders and the decay can be well treated by 3-body theory.

the 2nd paragraph describes the decay channel of 8C and 8B. it uses the Q-value to explain why the 2-protons decay is possible. it is because the 1-proton decay has negative binding energy, thus, it require external energy to make it decay. while 2-protons decay has positive binding energy, thus, the decay will automatic happen in order to bring the nucleus into lower energy state. it also consider the isospin, since the particle decay is govt by strong nuclear force, thus the isospin must be conserved. and this forbid of 1-proton decay.

it explains further on the concept of 2-protons decay and 2 1-protons decay. it argues that, in the 8C, the 2-protons decay is very short time, that is reflected on the large energy width, make the concept of 2 1-protons decay is a unmeasurable concept. however, for the 8Be, the life time is 7 zs (zepto-second 10^{-21} ), the 8Be moved 100 fm ( femto-meter $latex 10^{-15} ), and this length can be detected and separate the 4-protons emission in to 2 2-protons decay.

the 3rd paragraph explain the experiment apparatus – detector.

the 4th paragraph explains the excitation energy spectrum for 8C, 6Be.

the 5th and 6th paragraphs explain the excitation energy spectrum for the 6Be form 8C decay. since the 2 steps 2 – protons decay has 4 protons. the identification for the correct pair of the decay is important. they compare the energy spectrum for 8C , 6Be and 6Be from decay to do so.

the 7th paragraph tells that they anaylsis the system of 2-protons and the remaining daughter particle, by moving to center of mass frame ( actually is center of momentum frame ) and using Jacobi T coordinate system, to simplify the analysis. the Jacobi T coordinate is nothing but treating the 2-protons the 2 protons are on the arm of the T, and the daughter particle is on the foot of the T.