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An old campfire is uncovered during an archacological dig. Its charcoal is found to contain less than 1/1000 the normal amount ofC. Find the minimum age (in years) of the charcoal. Carbon-14 has a half-life of 5730 years which means half of it will be gone aftler 5730 years. You may find the equality 210- 1024 to be useful, Numeric : A numeric value is expected and not an expression. min

Sagot :

Answer:

57104 yrs old

Explanation:

Its charcoal is found to contain less than 1/1000 the normal amount ofC-14.

1/1000 = (1/2)^n     n = number of half lives  = 9.965 half lives

9.965 half lives * 5730 yrs/half life = 57,104 years

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