A tree sample was uprooted and buried 60,000 years ago during part of the Wiscon

Question

A tree sample was uprooted and buried 60,000 years ago during part of the Wisconsin glaciation. The tree contains 50 g of carbon when it is discovered.

A) If 1 in 1012 carbon atoms in a fresh tree sample are carbon-14, how many carbon-14 atoms would be in 50 g of carbon from a fresh tree?

-Answer: N0 = 2.5x10^12

B) Calculate the carbon-14 activity of the sample. Tthe half-life of carbon-14 is T=5700 years.

Express your answer using two significant figures.

C)Determine the age of the buried tree if its 50 g of carbon has an activity of 2.2 s1.

Express your answer using two significant figures.

Explanation / Answer

A)

we know that

moles = mass / molar mass

molar mass of Carbon = 12 g/mol

given

mass of Carbon = 50 g

so

moles of carbon = 50 / 12 = 4.16667

now

number of Carbon atoms = moles x avagadro number ( 6.022 x 10^23)

so

number of carbon atoms = 4.16667 x 6.022 x 10^23

number of carbon atoms = 2.509 x 10^24

now

1 in 10^12 carbon atoms are carbon-14

so

number of carbon-14 atoms = 2.509 x 10^24 / 10^12

number of carbon-14 atoms = 2.509 x 10^12

B)

now

the activity is given by

A = lamda x N

where

lamda = decay constant

now

for radioactive decay processes

lamda = ln2 / t1/2

given

half life = 5700 yr = 1.7976 x 10^11 s

so

lamda = ln2 / ( 1.7976 x 10^11)

lamda = 3.856 x 10-12

now

Activity = 3.856 x 10-12 x 2.509 x 10^12

activity = 9.6746

C)

lamda = ln2 / 5700

lamda = 1.216 x 10-4 yr -1

now

A= Ao x e^(-lamda x t)

2.2 = 9.6746 x e^(-1.216 x 10-4 x t )

t = 12179.659 yr

so

the age of the buried tree is 1.2 x 10^3 years

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