select all the correct answers.\nthe isotope radium - 226 decays into radon - 222, with a half - life of…

select all the correct answers.\nthe isotope radium - 226 decays into radon - 222, with a half - life of around 1,600 years. if a rock contained 6 grams of radium - 226 when it reached its closure temperature but only 0.375 grams when it was discovered, which two statements about the rock are true?\n□ the rock reached its closure temperature 6,400 years ago.\n□ the rock reached its closure temperature 4,800 years ago.\n□ the rock had 2.625 grams of radon - 222 1,600 years ago.\n□ when the rock was discovered, it had 5.625 grams of radon - 222.\n□ when the rock was discovered, it had 3.375 grams of radon - 222.

select all the correct answers.\nthe isotope radium - 226 decays into radon - 222, with a half - life of around 1,600 years. if a rock contained 6 grams of radium - 226 when it reached its closure temperature but only 0.375 grams when it was discovered, which two statements about the rock are true?\n□ the rock reached its closure temperature 6,400 years ago.\n□ the rock reached its closure temperature 4,800 years ago.\n□ the rock had 2.625 grams of radon - 222 1,600 years ago.\n□ when the rock was discovered, it had 5.625 grams of radon - 222.\n□ when the rock was discovered, it had 3.375 grams of radon - 222.

Answer

Answer:

  • The rock reached its closure temperature 4,800 years ago.
  • When the rock was discovered, it had 5.625 grams of radon - 222.

Explanation:

Step1: Determine the number of half - lives

The formula for radioactive decay is $N = N_0(\frac{1}{2})^n$, where $N$ is the final amount of the radioactive substance, $N_0$ is the initial amount, and $n$ is the number of half - lives. Given $N_0 = 6$ grams, $N=0.375$ grams. We solve for $n$: $0.375 = 6\times(\frac{1}{2})^n$. Then $(\frac{1}{2})^n=\frac{0.375}{6}=0.0625$. Since $0.0625=\frac{1}{16}=(\frac{1}{2})^4$, $n = 4$.

Step2: Calculate the time passed

The half - life of radium - 226 is $T_{1/2}=1600$ years. The time passed $t=n\times T_{1/2}$. Substituting $n = 4$ and $T_{1/2}=1600$ years, we get $t = 4\times1600=4800$ years.

Step3: Calculate the amount of radon - 222

The initial amount of radium - 226 was 6 grams and the final amount was 0.375 grams. The amount of radium - 226 that decayed is $6 - 0.375=5.625$ grams. Since each gram of radium - 226 that decays produces 1 gram of radon - 222, when the rock was discovered, the amount of radon - 222 was 5.625 grams. 1.600 years ago, the amount of radium - 226 was half of the initial amount (1 half - life), so it was 3 grams, and the amount of radon - 222 was $6 - 3=3$ grams, not 2.625 grams.