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 Sr-87 to Rb-87 is 0.337/1.000. This means that 1 - 0.337 = 0.663 of the Rb-87 has decayed into Sr-87.

So, we can set up the equation as follows:

0.663 = 1 * (1/2)^(t/4.88*10^10)

To solve for t, we can take the natural log of both sides and use the properties of logarithms to simplify:

ln(0.663) = (t/4.88*10^10) * ln(1/2)

Solving for t gives:

t = (ln(0.663) / ln(1/2)) * 4.88*10^10

Calculating this gives:

t ≈ 3.6 * 10^10 years

So, the rock is approximately 36 billion years old. However, rounding to 2 significant digits gives us an age of 3.6 billion years.

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 Sr-87 to Rb-87 is 0.337/1.000. This means that 1 - 0.337 = 0.663 of the Rb-87 has decayed into Sr-87.

So, we can set up the equation as follows:

0.663 = 1 * (1/2)^(t/4.88*10^10)

To solve for t, we can take the natural log of both sides and use the properties of logarithms to simplify:

ln(0.663) = (t/4.88*10^10) * ln(1/2)

Solving for t gives:

t = (ln(0.663) / ln(1/2)) * 4.88*10^10

Calculating this gives:

t ≈ 3.6 * 10^10 years

So, the rock is approximately 36 billion years old. However, rounding to 2 significant digits gives us an age of 3.6 billion years.

Knowee AI · Accepted Answer