To solve this problem, we need to use the formula for radioactive decay, which is:

N = N0 * (1/2)^(t/T)

where:
N is the final quantity of the substance,
N0 is the initial quantity of the substance,
t is the time that has passed, and
T is the half-life of the substance.

In this case, we know that the ratio of strontium-87 to rubidium-87 is 0.799, so we can say that N/N0 = 0.799.

We also know that the half-life T of rubidium-87 is 4.88 billion years.

Substituting these values into the formula, we get:

0.799 = (1/2)^(t/4.88)

To solve for t, we can take the natural logarithm of both sides:

ln(0.799) = (t/4.88) * ln(1/2)

Solving for t gives:

t = 4.88 * (ln(0.799) / ln(1/2))

Calculating this gives:

t ≈ 6.43 billion years

So, the age of the rock is approximately 6.43 billion years, rounded to two significant digits.

Question

To solve this problem, we need to use the formula for radioactive decay, which is:

N = N0 * (1/2)^(t/T)

where:
N is the final quantity of the substance,
N0 is the initial quantity of the substance,
t is the time that has passed, and
T is the half-life of the substance.

In this case, we know that the ratio of strontium-87 to rubidium-87 is 0.799, so we can say that N/N0 = 0.799.

We also know that the half-life T of rubidium-87 is 4.88 billion years.

Substituting these values into the formula, we get:

0.799 = (1/2)^(t/4.88)

To solve for t, we can take the natural logarithm of both sides:

ln(0.799) = (t/4.88) * ln(1/2)

Solving for t gives:

t = 4.88 * (ln(0.799) / ln(1/2))

Calculating this gives:

t ≈ 6.43 billion years

So, the age of the rock is approximately 6.43 billion years, rounded to two significant digits.

Knowee AI · Accepted Answer