contestada

The fossilized remains of a plant were found at a construction site. The fossilized remains contain 1/32 the amount of carbon-14 that is present in a living plant.

Determine the approximate age of these fossilized remains.

Respuesta :

pstnonsonjoku

Answer:

28645 years

Explanation:

Given the formula;

0.693/t1/2 = 2.303/t log (No/N)

Given that N = 1/32 No

Note;

t1/2 = half life of Carbon-14

t = time required for N amount of carbon -14 to remain= 5,730 years

No= amount of carbon 14 initially present

N = amount of carbon-14 after time t

Substituting values;

0.693/5,730 = 2.303/t log (No/1/32No)

0.693/5,730 = 2.303/t log 32

1.21 * 10^-4      = 3.466/t

t = 3.466/1.21 * 10^-4

t = 28645 years

Any preserved imprints, remains and traces of once a living organism from history are called fossils. By utilizing the radioactive property of organic compounds one can determine the age of an object.

The approximate age of the fossilizes remain is 28645 years.

This can be estimated as:

The formula used will be:

[tex]\dfrac{0.693}{\dfrac{t1}{2}}= \dfrac{2.303}{t} log (\dfrac{No}{N})[/tex]

Where,

  • Half-life of Carbon-14 =[tex]\dfrac{t1}{2}[/tex]  

  • Time required (t) for N amount of carbon -14 to remain= 5,730 years

  • Amount of carbon 14 initially present = [tex]N_{o}[/tex]

  • Amount of carbon-14 after time t = N

Given,

N = [tex]{\dfrac{1}{32} No[/tex]

Replacing values in formula:

[tex]\dfrac{0.693}{5,730} & = \dfrac{2.30}{t} log \:({{\dfrac{No}{\dfrac{1}{32} No}})[/tex]

[tex]\dfrac{0.693}{5,730} & = \dfrac{2.30}{t} \;log 32[/tex]

[tex]1.21 \times 10^{-4} = \dfrac{3.466}{t}[/tex]

[tex]t = \dfrac{3.466}{1.21} \times 10^{-4}[/tex]

[tex]t = 28645 \:\text{years}[/tex]

Therefore, the approximate age of the fossil is 28645 years.

To learn more about fossils and their age follow the link:

https://brainly.com/question/8917701

Otras preguntas