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A tree falls in a forest. How many years must pass before the $^{14} \mathrm{C}$ activity in 1.00 $\mathrm{g}$ of the tree's carbon drops to 1.00 decay per hour?

$5.63 \times 10^{4}$ years

Physics 103

Chapter 31

Radioactivity and Nuclear Physics

Nuclear Physics

Hope College

University of Winnipeg

McMaster University

Lectures

02:26

In physics, nuclear physic…

08:29

In physics, nuclear proper…

03:27

A piece of wood from an an…

03:44

Dating a Tree How long doe…

04:34

A piece of charcoal used f…

03:01

After a plant or animal di…

05:17

It takes 5700 years for an…

00:47

The carbon-14 decay rate o…

01:27

Involve exponential decay.…

02:46

A 5.00 g charcoal sample f…

04:39

Estimating the Age of a Tr…

01:30

05:53

01:00

Assume a constant $1^{14} …

00:45

A wooden object from the s…

04:28

A piece of charcoal is fou…

02:04

Carbon Dating The amount $…

01:32

SSM How many years are nee…

06:42

A 12.0-g sample of carbon …

03:09

$^{209}_{79}$ $\mathrm{P}o…

02:30

A radioactive element deca…

03:20

The half-life of titanium-…

the number of years past for the decay to reach a one day caper are that he's a time is it called Eleanor the number of items initially derided by number off the items final derided by the constant airborne 693 93 times in new miniter half life oflife we have 0.6 million trees a constant so we'll find and I so initial and I can be found by multiplying for carbon 40 1.3 turns into the poor were minus 12. Do there, uh, to the given mas divided by the molar mass times they're about those number of particles so given must be here for one gram Wonder I'm the mole a mosque Multiply by There I got his number of particles with December bringing here also, they were gonna number which is six forms there too. Dobbs dented powered dented power to 23. Our 23 gives us off the initial number off Peyton's. That is a cool toe. Six 0.52 times 10 to the power are 10 a DMZ. So this is 10 8 abs, 94 items. Ah, Where is the number of freedoms? Final number victims we have can be found by using relationship for a duty that is our times half life derided by 0.6 million. Three. So activity we have he's ah ah, one day cape are so when the Cape are in the second with today's of one over 3600 2 600 times 1/2 life is 1.8 one times 10 to the power 11 11 in the seconds derided by the constant 0.693 And this gives us an F l u E seven point to six times 10 to the power 78 times seven. Peyton's So we'll substitute this value for N F N and I in a poem equation. And where is that half life we have here? So, Ellen off N f and I and so an eye over an f will come here in this business of your life. Um, and half life we have in years the Aires 57 30 divided by the constant zero foreign 693 Did you find the expression we get here? The time, which is a 5162 times 10 to the power. Four years, four years. And that's the problem. Thank you for

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