Question

Ru-106 can be used to treat melanoma in the eye. This isotope decays by \beta emission with a half-life of 373 days. Suppose a fresh sample contains $6 \times 10^6$ nuclei of Ru-106. What is the decay rate of the sample after 700 days. 35 Bq 86 Bq 240 Bq 180 Bq 449 Bq 4 Bq

          Ru-106 can be used to treat melanoma in the eye. This isotope decays by \beta emission with a half-life of 373 days. Suppose a fresh sample contains $6 \times 10^6$ nuclei of Ru-106. What is the decay rate of the sample after 700 days.
35 Bq
86 Bq
240 Bq
180 Bq
449 Bq
4 Bq
        
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Ru-106 can be used to treat melanoma in the eye. This isotope decays by βemission with a half-life of 373 days. Suppose a fresh sample contains 6 × 10^6 nuclei of Ru-106. What is the decay rate of the sample after 700 days.
35 Bq
86 Bq
240 Bq
180 Bq
449 Bq
4 Bq

Added by Kathleen R.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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06 can be used to treat melanoma in the ye.This isotope decays by emission with a hat-fe o 373 days.Suppoe fresh mple conan ucii of Ru-106.What Is the decay rate of the sample after 700 days 35B 86 240 180Bq O449B O4Bq
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Transcript

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00:01 We know that the decay of a radioactive isotope follows an exponential decay model which can be expressed as nt is equal to n0 e to the power minus lambda t where nt is the number of radioactive nuclei at time t, n0 is the initial number of radioactive nuclei, lambda is the decay constant and e is the base of the natural logarithm.
00:25 We also know that the decay constant is related to the half -life by the equation lambda is equal to ln 2 by t half where ln is the natural logarithmic and t half is the half -life.
00:43 We can use the given data to find the decay constant lambda, let's call the initial number of radioactive nuclei n0 then we have n0 e to the power minus lambda t1 is equal to 6833 and n0 e to the power minus lambda t2 is equal to 3530 where t1 is the initial time and t2 is the time 4 .88 days later.
01:18 Dividing the second equation by the first equation we get e to the power minus lambda t2 minus t1 is equal to 3530 divided by 6833.
01:35 Taking the natural logarithmic both sides we get lambda t2 minus t1 is equal to ln 3530 divided by 6833.
01:49 Solving for lambda we get lambda is equal to ln 3530 divided by 6833 divided by 822 minus t1...
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