NEET-XII-Physics
46: The Nucleus
- #24The count rate from a radioactive sample falls from 4.0 × 106 per second to 1.0 × 106 per second in 20 hours. What will be the count rate 100 hours after the beginning?Ans : Given:
Initial count rate of radioactive sample, A0 = 4 × 106 disintegration/sec
Count rate of radioactive sample after 20 hours, A = 1 × 106 disintegration/sec
Time, t = 20 hours
Activity of radioactive sample `` \left(A\right)`` is given by
`` A=\frac{{A}_{0}}{{2}^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}}``
`` \,\mathrm{\,Here\,},{T}_{1/2}=\,\mathrm{\,Half\,}-\,\mathrm{\,life\,}\,\mathrm{\,period\,}``
`` \,\mathrm{\,On\,}\,\mathrm{\,substituting\,}\,\mathrm{\,the\,}\,\mathrm{\,values\,}\,\mathrm{\,of\,}{A}_{0}\,\mathrm{\,and\,}A,\,\mathrm{\,we\,}\,\mathrm{\,have\,}``
`` {2}^{\frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}={2}^{2}``
`` \Rightarrow \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2``
`` \Rightarrow \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=t/2=20\,\mathrm{\,h\,}/2=10\,\mathrm{\,h\,}``
`` ``
100 hours after the beginning,
`` \,\mathrm{\,Count\,}\,\mathrm{\,rate\,},A"=\frac{{A}_{0}}{{2}^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}}``
`` \Rightarrow A"=\frac{4\times {10}^{6}}{{2}^{100/10}}``
`` =0.390625\times {10}^{4}``
`` =3.9\times {10}^{3}\,\mathrm{\,disintegrations\,}/\,\mathrm{\,sec\,}``
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