The chart displays the rate of radioactive decay for iodine-131.

Question

Decay of lodine-131

amount (percent) 688888888

24 6 8 10 12 14 16 18 20 22:24 26 28 30 32 34

time (days)

According to the graph, what is the period of half-Sife for iodine-131?

(1 point)

24 days

16 days

8 days

4 days

Answer 1

The half-life of a radioactive substance is the time it takes for half of the original amount to decay. From the information provided, it appears we are working with a decay chart for iodine-131.

To determine the half-life, you would look for the point on the graph where the amount of iodine-131 has decreased to 50% of its original amount. If the graph showed, for example, that the amount of iodine-131 decreased from 100% to 50% over a specific time period, that time frame would indicate the half-life.

Since the specific values of the graph are not visible in your message, I cannot analyze it directly. However, if the options for the half-life are given as:

• 24 days
• 16 days
• 8 days
• 4 days

You would select the one that most closely corresponds to the amount decaying to 50% on the graph. If you find that it takes, for example, 8 days to go from 100% to 50%, then the half-life would be 8 days.

If you have more details or can describe the specific point on the chart where the amount reaches half of its original level, I'd be glad to assist you further!

Answer 2

What conclusion can be inferred from a 3.0 picogram (pg) sample of curium-245 with a half-life of 8500 years? (1 point)

The age of the curium-245 sample must be less than 8500 years.

After 8500 years, the curium-245 sample is expected to have a mass of approximately 6.0 picograms.

After 8500 years, the curium-245 sample is anticipated to possess a mass of roughly 1.5 picograms.

The age of the curium-245 sample must be at least 8500 years.

Answer 3

To analyze the given options, we need to understand how the half-life of a radioactive substance works. The half-life is the time it takes for half of the original sample to decay.

In this case, the half-life of curium-245 is 8500 years.

1. If you start with a 3.0 picogram (pg) sample of curium-245:
   • After 8500 years, half of that amount will remain: \[ \text{Remaining mass} = \frac{3.0 \text{ pg}}{2} = 1.5 \text{ pg} \]

Now we can analyze the options provided:

1. The age of the curium-245 sample must be less than 8500 years.

   • This option can’t be definitively concluded from the available information because we don’t know how long the sample has already been decaying.

2. After 8500 years, the curium-245 sample is expected to have a mass of approximately 6.0 picograms.

   • This is incorrect, since after 8500 years, the mass would be 1.5 picograms.

3. After 8500 years, the curium-245 sample is anticipated to possess a mass of roughly 1.5 picograms.

   • This is correct, as we calculated.

4. The age of the curium-245 sample must be at least 8500 years.

   • This cannot be concluded since the age could be less than 8500 years (if it has been recently measured) or more than 8500 years (if it has already gone through one or more half-lives).

Based on this analysis, the best conclusion is:

After 8500 years, the curium-245 sample is anticipated to possess a mass of roughly 1.5 picograms.

Answer 4

The table illustrates the decay in mass of a 59-gram sample of Bismuth-212 as time progresses

Time (seconds)

Bismuth-212 (grams)

0

60.5

121

59

29.5
x

After 121 seconds, how many grams of the sample will remain if Bismuth-212 has a half-life of 60.5 seconds?

(1 point)

3.69 g

14.75 g

7.38 g

29.5 g

Answer 5

To determine how much of the Bismuth-212 sample remains after 121 seconds, given its half-life of 60.5 seconds, we can follow these steps:

1. Determine how many half-lives have occurred in 121 seconds: \[ \text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} = \frac{121 \text{ seconds}}{60.5 \text{ seconds}} \approx 2 \] This means approximately 2 half-lives have passed.

2. Calculate the remaining mass after each half-life.

   • After 1st half-life (60.5 seconds): \[ \text{Remaining mass} = \frac{59 \text{ grams}}{2} = 29.5 \text{ grams} \]

   • After 2nd half-life (121 seconds): \[ \text{Remaining mass} = \frac{29.5 \text{ grams}}{2} = 14.75 \text{ grams} \]

3. Conclusion: After 121 seconds, the remaining mass of the Bismuth-212 sample is 14.75 grams.

Thus, the correct answer is:

14.75 g.