The model for radioactive decay is y=y0e−kt. A radioactive substance has a half-life of 1150 years. If 15 grams are present today, in how many years will 10 grams be present? (Round the value of k to 7 decimal places.) Round your answer to two decimal places, if necessary.

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Pohanginah · Accepted Answer

Step 1  Given: The exponential equation y=y0e−kt, half life =1150 years.  To find: If 15 gm. are present today, in how many years will 10 grams be present.  Concept Used: The above equation can be solved by making equation and using elimination method solve accordingly.  Step 2  Explanation- Rewrite the given equations,  y=y0e−kt⋯⋯(1)  As the half life of radioactive element is 1150 years, so we can write as y=y02 and t=1150 years, so from the above expressions, we can write as,  y02=y0e−k⋅1150  0.5=e−1150k  Now, taking loge both sides, we get,  ln⁡e−1150k=ln⁡(0.5)  −1150k=−0.693  k=0.6931150  k=0.0006027  Step 3  Now, substituting k=0.0006027 in the equation (1), we get,  y=y0e−0.0006027t⋯⋯(2)  As we have to find the time in years in which the amount of radioactive element is 10 grams present.   So, substituting y0=15gm,y=10gm in the equation (2), we get,  10=15e−0.0006027t  1015=e−0.0006027t  0.67=e−0.0006027t  Now, taking loge both sides, we get,  −0.0006027t=ln⁡(0.67)  t=−0.40050.0006027  t=664.47 years  So, in 664.47 years, the radioactive element will remain only 10 gram.   Answer- Hence, in 664.47 years, the radioactive element will remain only 10 gram.

The model for radioactive decay is y=y0e^{-kt}. A radioactive substance ha

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