Select the correct answer from the drop-down menu.

A scientist digs up a sample of Arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 grams of krypton-81.

If the half-life of krypton-81 is 229,000 years, how much krypton-81 was present when the ice first formed?
Use the formula [tex]N=N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}[/tex].

The ice originally contained
_____ grams of krypton-81.



Answer :

To solve this problem, we need to use the concept of radioactive decay and the given formula. The key is to determine how much krypton-81 was present initially when the ice was formed, based on its current amount and its half-life.

Here's the detailed step-by-step solution:

1. Identify the Given Values:
- Age of the ice = 458,000 years
- Current amount of krypton-81 = 1.675 grams
- Half-life of krypton-81 = 229,000 years

2. Understand the Formula:
The formula for radioactive decay is:
[tex]\[ N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \][/tex]
where:
- [tex]\(N\)[/tex] is the remaining amount of substance (krypton-81 in this case).
- [tex]\(N_0\)[/tex] is the initial amount of the substance.
- [tex]\(t\)[/tex] is the elapsed time (age of the ice).
- [tex]\(T\)[/tex] is the half-life of the substance.

3. Rearrange the Formula:
To find [tex]\(N_0\)[/tex], we rewrite the formula:
[tex]\[ N_0 = \frac{N}{\left(\frac{1}{2}\right)^{\frac{t}{T}}} \][/tex]

4. Substitute the Known Values:
Plug in the known values into the rearranged formula:
[tex]\[ N_0 = \frac{1.675}{\left(\frac{1}{2}\right)^{\frac{458,000}{229,000}}} \][/tex]

5. Simplify the Exponent:
Simplify the exponent in the denominator:
[tex]\[ N_0 = \frac{1.675}{\left(\frac{1}{2}\right)^2} = \frac{1.675}{\frac{1}{4}} = 1.675 \times 4 \][/tex]

6. Calculate the Initial Amount:
[tex]\[ N_0 = 6.7 \text{ grams} \][/tex]

So, the ice originally contained 6.7 grams of krypton-81. Choose "6.7" in the drop-down menu.

Congratulations, we have completed the step-by-step solution!