ln(M/Mo) = -kt
ln = natural log
M = mass left (not needed)
Mo = original mass (not needed)
(M/Mo) = 0.809
k = 0.693/T(half)
T(half) = 5.73X10^3 years.
linen scrolls contain 80.9% of Carbon-14. how old are the scrolls if the half-life for deacy is 5.73X10^3??
plz tell me what type of equation to use.
3 answers
ln(M/Mo) = -kt
ln = natural log
M = mass left (not needed)
Mo = original mass (not needed)
(M/Mo) = 0.809
k = 0.693/T(half)
T(half) = 5.73X10^3 years.
ln = natural log
M = mass left (not needed)
Mo = original mass (not needed)
(M/Mo) = 0.809
k = 0.693/T(half)
T(half) = 5.73X10^3 years.
ln(M/Mo) = -kt
ln = natural log
M = mass left (not needed)
Mo = original mass (not needed)
(M/Mo) = 0.809
k = 0.693/T(half)
T(half) = 5.73X10^3 years.
ln = natural log
M = mass left (not needed)
Mo = original mass (not needed)
(M/Mo) = 0.809
k = 0.693/T(half)
T(half) = 5.73X10^3 years.