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qcm concours 2011


Khaoula38
Go to solution Solved by Claugomes3,

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  • Solution

Bonsoir 😄

 

En fait, il faut juste transposer les infos de l'énoncé sous forme de calcul, on sait que l'activité de la source est de 200x10^6 Bq et qu'il y a 10% de Be qui se désintègrent pour former l'élément X donc 200x10^6 x 10/100 = 20x10^6 éléments. Ces éléments vont ensuite libérer leur excès d'énergie (0,480 MeV) sous forme de photons gamma pour retrouver la stabilité sous forme de Li. 

Donc, on aura bien 20x10^6 photons gamma de 480 keV.

 

J'espère que ce sois plus clair.

 

Bonne soirée! 😊

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