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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 :

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Answer:

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Step-by-step explanation:

Mass of  samarium-151  originally present (No) = 5 g

Mass of  samarium-151 discovered (N) = 0.625 grams

time taken = ?

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

From

N/No = (1/2)^t/t1/2

0.625/5 = (1/2)^t/96

0.125 = (1/2)^t/96

1/8 = (1/2)^t/96

(1/2)^3 = (1/2)^t/96

3 = t/96

t = 3 * 96

t = 288 years

The mass of europium-151 when the rock was discovered is obtained from;

Fraction of  samarium-151  left = 1/8

Fraction of  europium-151 formed = 1 - 1/8 = 7/8

Mass of  europium-151 = 7/8 * 5 = 4.375 g

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