After 148 days, how many curies would be left from an original 98 Ci of Ir-192?

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Multiple Choice

After 148 days, how many curies would be left from an original 98 Ci of Ir-192?

Explanation:
To determine how many curies of Iridium-192 (Ir-192) would remain after 148 days from an initial quantity of 98 curies, it is essential to understand the half-life of the isotope in question. The half-life of Ir-192 is approximately 74 days. This means that every 74 days, the amount of Ir-192 will reduce to half of its previous amount. In the context of the question, after 148 days, we identify how many half-lives have elapsed. Since 148 days is approximately two half-lives (148 days ÷ 74 days/half-life = 2), we can calculate the remaining activity as follows: 1. After the first half-life of 74 days, the amount would be halved: 98 Ci ÷ 2 = 49 Ci. 2. After the second half-life of another 74 days (totaling 148 days), the amount would again be halved: 49 Ci ÷ 2 = 24.5 Ci. Therefore, after 148 days, 24.5 curies of Ir-192 would remain. This calculation aligns perfectly with the concept of radioactive decay and half-lives, confirming that

To determine how many curies of Iridium-192 (Ir-192) would remain after 148 days from an initial quantity of 98 curies, it is essential to understand the half-life of the isotope in question. The half-life of Ir-192 is approximately 74 days. This means that every 74 days, the amount of Ir-192 will reduce to half of its previous amount.

In the context of the question, after 148 days, we identify how many half-lives have elapsed. Since 148 days is approximately two half-lives (148 days ÷ 74 days/half-life = 2), we can calculate the remaining activity as follows:

  1. After the first half-life of 74 days, the amount would be halved:

98 Ci ÷ 2 = 49 Ci.

  1. After the second half-life of another 74 days (totaling 148 days), the amount would again be halved:

49 Ci ÷ 2 = 24.5 Ci.

Therefore, after 148 days, 24.5 curies of Ir-192 would remain. This calculation aligns perfectly with the concept of radioactive decay and half-lives, confirming that

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