Question

The isotope samarium-151 decays into europium-151, with a half-life of around 96.6 years. A rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered. The time since the rock reached its closure temperature is ( ) years. When the rock was discovered, it had ( ) grams of europium-151.

A) t≈96.6 years; 0.6250.625 grams
B) t≈193.2 years; 1.251.25 grams
C) t≈289.8 years; 0.31250.3125 grams
D) t≈96.6 years; 1.251.25 grams

Answer 1

The rock is approximately 442.4 years old and had 1.25 grams of europium-151 when discovered.

Step-by-step explanation:

To determine the age of the rock, we can use the concept of radioactive decay and the half-life of samarium-151. The half-life is the time it takes for half of a radioactive substance to decay. In this case, the half-life of samarium-151 is approximately 96.6 years. By calculating the number of half-lives that have passed, we can determine the age of the rock.

First, we need to find the number of half-lives. The initial mass of samarium-151 is 5 grams, and the final mass is 0.625 grams. Each half-life reduces the mass by half, so the number of half-lives is given by:

Number of half-lives = log2(initial mass / final mass)

Number of half-lives = log2(5 / 0.625) ≈ 4.585

Since each half-life is approximately 96.6 years, the age of the rock is approximately 96.6 years x 4.585 ≈ 442.4 years.

When the rock was discovered, the mass of europium-151 can be calculated using the mass of samarium-151 and the number of half-lives. Each decay of samarium-151 produces one atom of europium-151. Therefore, the mass of europium-151 is approximately 0.625 grams x 24.585 ≈ 1.25 grams.