High School

Type the correct answer in each box. Use numerals instead of words. Express your answer as a decimal number.

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.

Answer :

When the rock was discovered, it had approximately 4.375 grams of europium-151.

To determine the time since the rock reached its closure temperature, we can use the concept of radioactive decay and the relationship between the half-life and the amount of remaining substance.

The equation for radioactive decay is given by:

N(t) = N₀ * (1/2)^(t / t₁/₂)

Where N(t) represents the remaining amount of substance at time t, N₀ is the initial amount of substance, t₁/₂ is the half-life, and t is the time that has elapsed.

Initial amount of samarium-151 (N₀) = 5 grams

Remaining amount of samarium-151 (N(t)) = 0.625 grams

Half-life of samarium-151 (t₁/₂) = 96.6 years

We can rearrange the equation to solve for time (t):

t = (t₁/₂) * log(N(t) / N₀) / log(1/2)

Substituting the values into the equation:

t = (96.6 years) * log(0.625 grams / 5 grams) / log(1/2)

Calculating this expression:

t ≈ (96.6 years) * (-0.3010) / (-0.6931) ≈ 41.93 years

Therefore, the time since the rock reached its closure temperature is approximately 41.93 years.

To calculate the grams of europium-151 when the rock was discovered, we can use the fact that europium-151 is the decay product of samarium-151. Since the remaining amount of samarium-151 is 0.625 grams, the grams of europium-151 will be the difference between the initial amount and the remaining amount of samarium-151:

Grams of europium-151 = 5 grams - 0.625 grams = 4.375 grams

For more such question on europium visit:

https://brainly.com/question/11367704

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