Isospin Symmetry test on the semimagic 44Cr Toward the dripline in the f7/2 shell I would like to propose the study of 44Cr which is a semimagic nucleus with the valence protons in the f7/2 orbital

44Cr N=20, 40Ca +4 protons N=Z Mid mass, Tz=-2 36Ca@RISING This is the nucleus: It stays in the N=20 and has a 4 protons excess with respect to 40Ca. It is one of the few cases where we have the possibility of studying a medium mass nucleus with an isospin value of Tz=-2. During the fast beam campaign of RISING an other Tz = -2 has been studied: 36Ca That experiment is a good reference and I will use it for the experimental part of the proposal.

T=1 and T=2 mirror nuclei Shell evolution Gaps Z=14 and N=14 No cross shell excitations The data from that experiment have given information on the shell evolution. In this figure you can see how the dashed line obtained after modifying the gaps Z=14 and N=14 of the the USD interaction reproduces the systematics of the MED within the full SD shell. So, 36Ca nucleus has been interpreted with a shell model calculation restricted to the sd shell, without considering any particle-hole excitation across the Z,N=20 f7/2 f7/2 N,Z=20 N,Z=20 d3/2 d3/2 s1/2 s1/2 d5/2 d5/2 π ν π ν 36Ca 36S

Cross conjugate nuclei f7/2 f7/2 N,Z=20 N,Z=20 d3/2 d3/2 s1/2 s1/2 d5/2 d5/2 π ν π ν 44Cr 44Ca 44Cr is the natural continuation of what has been done during the previous campaign: The 2 mirrors 44Cr and 44Ca are cross conjugate of the 36Ca and 36S because they have 4 particles instead of 4 holes on the 40Ca. It would be interesting to see for example a correlation in between the MEDs and using the information from 44Cr nucleus we could reconsider some calculation done for the A=36 T=2 mirrors. f7/2 f7/2 N,Z=20 N,Z=20 d3/2 d3/2 s1/2 s1/2 d5/2 d5/2 π ν π ν 36Ca 36S

44Cr 44Cr 44Ca 44Ca Exp kb3g gxpf1a B(E2)[e2fm4] 104 9.8 10.3 f5/2 N,Z=20 N,Z=20 π ν π ν 44Cr 44Ca 1360 2+ 1157 2+ 1248 2+ For example Silvia Lenzi using Antoine calculated the level scheme for the 44Ca and can reproduce it. Here you see results of a calculation performed taking a shell closure at N,Z=20 as was done for the calculations of 36Ca. This calculation reproduces the excitation energy but not at all the B(E2) of 44Ca that is the mirror of 44Cr… 0+ 0+ 0+ 44Ca Exp kb3g gxpf1a B(E2)[e2fm4] 104 9.8 10.3

44Cr 44Cr 44Ca 44Ca Exp kb3g gxpf1a sdfp B(E2)[e2fm4] 104 9.8 10.3 N,Z=20 N,Z=20 d3/2 d3/2 s1/2 s1/2 π ν π ν 44Cr 44Ca 1571 2+ 1360 2+ 1157 2+ 1248 2+ … instead, if the valence space is extended in order to include possible particle-hole excitations, this is the result: the experimental data are well reproduced and in particular the B(E2) has the correct value, accounting for the collectivity coming from the mixing of closed shell states with cross shell particle-hole excitations. 0+ 0+ 0+ 0+ 44Ca Exp kb3g gxpf1a sdfp B(E2)[e2fm4] 104 9.8 10.3 105.6

Particle-hole cross-shell excitations 1571 2+ 1360 2+ 1157 2+ 1248 2+ It has been shown by Caurier that the sdfp interaction is able to reproduce 2+, B(E2) and also 3- in the f7/2 region. We decided to propose the study of 44Cr because our collaboration has experience in these studies and on the theoretical description of MED in terms of shell model calculations 0+ 0+ 0+ 0+ 44Ca Exp kb3g gxpf1a sdfp B(E2)[e2fm4] 104 9.8 10.3 105.6

44Cr Isospin symmetric: 44Ca 44Cr 44Ca 3307 3- 3285 6+ Sp = 2800 keV (SY) 2283 4+ The observables to be used in the study are the MED which have been shown to be really meaningful for the understanding of the changing of nuclear structure as a function of the angular momentum. Here, given the 2.8 MeV of separation energy for the protons, we expect to be able to see the excitation energies of the states till 6+ and also 3- in the 44Cr. 1157 2+ 0+ 0+ 44Cr 44Ca

f7/2 shell and INC nuclear forces 25Mn24 24Cr25 49 VCm, VCM VB The kind of information that we could get is exemplified from a previous study done on f7/2 nuclei. For example in A=49 mirror nuclei the symmetry breaking has been studied in detail through the dependence of the MED on the angular momentum. Here, in the case of the A=49 mirror nuclei we can see the goodness of the theoretical description and the different contributions that enter the calculations. Three contributions have to be added in order to reproduce the experimental data: The Coulomb monopole term, The Coulomb Multipole term And also a third term breaking the isospin symmetry has to be added. This last term is not explained by Coulomb effects and has been suggested to be of nuclear origin. From the study of these MED important information has been obtained regarding: How the nucleus generates its angular momentum The evolution of the deformation along a rotational band Isospin non-conserving terms in the nuclear interaction and the configuration of the states It is important to note that the good description of the MED data here observed, has been also achieved for all the mirror partners studied in the f7/2 shell, without changing any parameter. In the case of A=44 we will be limited to 6+ level but we will study systematics together with other nuclei From the MED we extract information of nuclear structure properties How the nucleus generates its angular momentum Evolution of the deformation along a rotational band Isospin non-conserving terms in the nuclear interaction Learn about the configuration of the states

36Ca from 40Ca Fragmentation: 40Ca → 37Ca 38 µbarn Knock-out: 37Ca → 36Ca 2 mbarn This is the reaction used to produce the 36Ca nucleus, a fragmentation of 40Ca and a knock out on the secondary target.

44Cr from 50Cr Fragmentation: 50Cr → 45Cr 1.5 µbarn Knock-out: 45Cr → 44Cr 2 mbarn The production of 44Cr is more difficult than 36Ca but we have also an improved setup. To produce the 44Cr we could use a 50Cr beam. This means that we have to remove 6 neutrons from the target, 5 in the first reaction and the last one at the secondary target. This gives a reduction in cross section with respect to the Ca case of a factor 20. But there is a better solution…

44Cr from 58Ni Fragmentation: 58Ni → 45Cr 0.6 µbarn Knock-out: 45Cr → 44Cr 2 mbarn … instead of using a 50Cr we could use a 58Ni. Here we have to remove many nucleons but the cross section scales down by a factor of 3 only while the primary beam intensity is a factor 20 higher.

Feasibility: fragmentation + knock-out Comparison to 36Ca: Cross section: /50 AGATA efficiency: x5 AGATA resolution: x2 Energy of the gamma (3.0→ 1.2 MeV): x3 Beam current: x30 (3*108→1010) 44Cr now is a factor 20 easier! To understand the feasibility here are the spectra published for the 36Ca. We have a reduction of 50 in cross section that is counter-balanced by AGATA efficiency and resolution, by the efficiency gain due to the lower energy of the gamma ray and by the higher beam current available for the 58Ni primary beam. This means that the experiment is expected to give a factor 3 more statistic and a factor of 2 better resolution This estimation has to be refined with LISE and MOCADI simulations. I will work on this together with Stephane and Cesar. P. Doornenbal et al. Physics Letters B 647 (2007) 237–242

Feasibility: fragmentation + coulex Directly produced 44Cr: 24 nbarn More than an order of magnitude lost Coulex on secondary target 2+ predicted collective (2p2h, 4p4h) Enhanced B(E2) Good excitation cross section (~200 mbarn) Factor of ~10 in statistics for the 2+ If it will turn out that the cross section is too small – I don’t think so – we have an other possibility. We could select a 44Cr secondary beam and perform a Coulex reaction. The production in this case would be higher because the B(E2) of the 0+ to 2+ excitation is really high and we would measure it. This would allow to measure only 2+ excitation energy and not 4 or 6+ giving a higher statistics by one order of magnitude.

WILKINSON - ISOSPIN IN NUCLEAR PHYSICS B(E1) in T=2 Janecke: EC (A,T,TZ)= EC0(A,T) - Tz EC1(A,T) + (3Tz2-T(T+1))EC2(A,T) Warburton “Corresponding E1 transition in conjugate nuclei have equal strength” T=2 Tz=-2 EC = EC0 + 2 EC1(A,2) + 6EC2(A,T) T=2 Tz=-1 EC = EC0 + 1 EC1(A,2) - 3EC2(A,T) T=2 Tz=0 EC = EC0 + - 6EC2(A,T) T=2 Tz=+1 T=2 Tz=+2 EC = EC0 - 2 EC1(A,2) + 6EC2(A,T) WILKINSON - ISOSPIN IN NUCLEAR PHYSICS