Calculate the half-life time of a radioactive isotope given that the initial count rate was 890 Bq & after 180 min. the count rate was found to be 750 Bq

t _{½ }= 0.693 × t ÷ ln A_{0}/A

t_{1/2 }= 0.693 × 180 ÷ ln (890 / 750)

t_{1/2} = 728.84 min

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t

_{½ }= 0.693 × t ÷ ln A_{0}/At

_{1/2 }= 0.693 × 180 ÷ ln (890 / 750)t

need an explanation for this answer? contact us directly to get an explanation for this answer_{1/2}= 728.84 min