Comments on OPERA OPERA is going on. We will have good Tau neutrino interactions within years, if νμ － ντ oscillation is the real solution. New Techniques developed in OPERA Very fast Read out system SUTS, Emulsion film with new AgBr crystals Low energy electron detection

MeV Energy electron signal in OPERA CS1-CS2 alignment using Compton electrons. Very good track connection between two emulsion layers σ ～ 1.8μm By S. Miyamoto ΔｙΔｙ ΔｙΔｙ

Double Beta decay detection by Nuclear Emulsion Neutrino less Double Beta Decay 1) Two electrons emitted from one point.  Micron accuracy measurement Chance coincidence rejection by the position measurement 2) Total energy  Tracking Calorimeter: Track length + Grain information (Z,A)→(Z+2,A) + e e ν e1 + ν e2 (Z,A)→(Z+2,A) + e e 2 - Emulsion Source

Set-up example Basic Element : ECC configuration Source layer+ Emulsion layers ( 100 Mo foil for example) Source (Foil) 10micron 50 micron emulsion layer on both sides of the foil. 1 Unit

Set-up example NEMO3 level Next generation Source mass 10kg 100kg Emulsion Amount 10 litter 100 litter (FNAL E531 level) (CHORUS level) Scanning load 200 days OPERA SUTS Need X10 faster system ! 20cm 25cm 50cm 45cm 2500 unit 4500 unit

Main Background : 214 Bi, 208 Tl (Max 3.17MeV beta) comes from Rn or contained U, Th chains. Those can be rejected by the existence of Alpha particles  easy in Emulsion About Background NEMO3

理科年表（丸善） 10μm βray α ray track. Uranium Chain Thorium Chain α decay βdecay Chain Products(Alpha and Beta) are contained within several micron^3

Summary OPERA is going on. R&D in OPERA  Applications (Double Beta, Directional Dark Matter detection etc) Compact Double beta experiment with nuclear emulsion is possible. Current main background from Th ・ U chain can be rejected by detecting the existence of αrays.

In the case of Emulsion.. In the emulsion, background caused by Th ・ U chain can be rejected by α ray track. Βray of Th ・ U chain exist with α ray! 10μm βray

Needed Mass of emulsion 100 Mo Status strongest limit T 1/2 0ν > year MoO 3 sheet(10μm thickness) enrichment of 100 Mo 90% ⇒ more than 1,000 kg emulsion (efficiency ～ 100%, 1year exp.) 150 Nd Status strongest limit T 1/2 0ν > year NdO 3 (10μm thickness) enrichment of 150 Nd 91% ⇒ more than 1kg emulsion (efficiency ～ 100%, 1year exp.) 96 Zr Status strongest limit T 1/2 0ν > year ZrO 2 (10μm thickness) enrichment of 96 Zr 57% ⇒ more than 100 kg emulsion (efficiency ～ 100%, 1year exp.) Status Nuclear Emulsion experiment Emulsion mass : 30,000kg Source material should be decided by relation between Q-value and background or compatibility with emulsion. For example

energy resolution At least, ⊿ E is about 20% by effective range. effective range By trace of the one track, ⊿ E is expected about 10% By developing the new emulsion of high sensitivity and having high AgBr density ( ex. Fine grain emulsion), pick up the some energy loss. (ionization, knock-on electron, Bremsstrahlung etc) ⇒ aim to 5%

Low Energy electron signal CS1-CS2 alignment in OPERA( Compton Alignment) By this technique, It is possible to detect the MeV electrons. σΔx ～ 2.1μm σΔy ～ 1.8μm By S. Miyamoto

2νββ and 0νββ p p e-e- ν n n e-e- ν W+W+ W+W+ n n e-e- W+W+ W+W+ e-e- p p ν R =ν R (T 1/2 2ν ) -1 =G 2ν (Q,Z) |M 2ν | 2 (T 1/2 0ν ) -1 =G 0ν (Q,Z) |M 0ν | 2 2 Which is neutrino Majorana or Dirac particle? How much is neutrino’s mass? Investigation of Lepton Flavor Violation Neutrino must be Majorana particle. ( ν = ν ） 2νββ0νββ

Enegy spectrum of double beta decay (Z,A)→(Z+2,A) + e e ν e1 + ν e2 (Z,A)→(Z+2,A) + e e 2 - Total energy of 2 electron is Q-value region of energy spectrum of 2ν mode electrons. S(0ν)/N(2ν) Energy resolution [%] The case of T 1/2 0νββ T 1/2 2νββ ～ 10 6 Energy resolution is very important to distinguish between 0νββ and 2νββ !!