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एक रेडियोएक्टिव पदार्थ कि सक्रियता 25 दिन में घटकर प्रारंभिक मन का `1//32` रह जाती है । रेडिएक्टिव पदर कि अर्द्ध-आयु कि गणना किजिय ।

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यदि किसी रेडियोएक्टिव पदर कि प्रारम्भ में सक्रियता `R_0` है, तब N अर्द्ध-आयुओ के पश्चात् बचे पदार्थ कि सक्रियता
`R=R_0(1/2)^n" ":." "R/R_0=(1/2)^n`
बचा पदार्थ अपने प्रारम्भिक मान का `1/32` रह जाता है, अर्थात `R/R_0=1/32=(1/2)^5`
अथवा `(1/2)^n=(1/2)^5" ":." "n=5`
अथवा पदार्थ कि अर्द्ध-आयु=विघटन का समय / अर्द्ध - आयुओ कि संख्या `=("25 दिन")/5= 5` दिन |

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