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Physique Noyau


jaimelespates
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Salut @jaimelespates

 

Calculons d'abord l'activité initiale du 99mTc :

A = \frac{ln2.m.Na}{M.T}=\frac{0,7.36.10^{-9}.6.10^{23}}{100.6.3600}

On a arrondi 99 à 100.

On simplifie les 6 et les 36 ce qui nous laisse :

A =\frac{0,7.10^{-9}.10^{23}}{10^{4}}=0,7.10^{10}Bq

Soit A = 7000 MBq.

 

Puis, on calcule l'activité après 12h = 2T.

A (t = n.T) = \frac{Ai}{2^{n}}

Soit : A (t = 2.T) = \frac{7000 MBq}{2^{2}} = 1750 MBq

Donc A faux et B vrai.

 

Convertissons ensuite en Ci.

1 Bq = 27.10^-12 Ci

175.10^7 Bq = x Ci

x = 27.10^-12 x 175.10^7 = 4625.10^-5 Ci = 46,25 mCi.

Donc C vrai.

 

D faux et E faux, la preuve on l'a calculé.

E serait vrai si 10 x 6h = 60h étaient passées.

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