About radioactivity

Thank you for your previous help.

I have done FLUKA radioactivity analysis,the result of which is given above

there are two vale 11-Na-24
i) 4.60E+12
ii) 3.43E+12 (isomers)

I have following question
i) Does it means that out of 4.60E+12 11-Na-24 3.43E+12 are in isomeric
state,so that when isomeric state will return to ground state it will
decay by ground state mode of 11-Na-24
    (since there is 99.9% chances that isomeric state of 11-Na-24 will go
directly to ground state)
ii) Also when in radioactivity card in FLUKA if I will keep the option of
patch isomers off,will I get aprroximately 4.60E+12 number of 11-Na-24.

Irradiation time: 86400. s

   Decay time : 0. s

   Detector n: 1 outsod0s
      (Region 6 Volume: 1. cmc,
       distr. type : 3 ,
       Z_max: 15, N-Z_max: 7, N-Z_min: -4)

      Tot. response (Bq/cmc) 8.03557972E+12 +/- 99. %
      ( --> Bq 8.03557972E+12 +/- 99. % )

   **** Residual nuclei distribution ****
   **** (Bq/cmc) ****

   A \ Z 5 6 7 8 9 10
  11 12 13 14 15
   24 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00
4.60E+12 0.00E+00 0.00E+00 0.00E+00 0.00E+00
       +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 %
+/-99.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 %
   23 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 1.06E+09
0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00
       +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/-99.0 %
+/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 %
   22 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00
2.42E+07 0.00E+00 0.00E+00 0.00E+00 0.00E+00
       +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 %
+/-99.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 % +/- 0.0 %

   A \ Z 1 2 3 4

   **** Isomers (Bq/cmc) ****
       A Z mth
        24 11 1 3.43E+12 +/- 99.0 %
Received on Fri May 11 2012 - 11:27:09 CEST