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PhysicsCollege Physics1st Edition

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Chapter 31, Problem 55PE
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College Physics

1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000

Solutions

Chapter
Section
Problem 1CQ
Problem 2CQ
Problem 3CQ
Problem 4CQ
Problem 5CQ
Problem 6CQ
Problem 7CQ
Problem 8CQ
Problem 9CQ
Problem 10CQ
Problem 11CQ
Problem 12CQ
Problem 13CQ
Problem 14CQ
Problem 15CQ
Problem 16CQ
Problem 17CQ
Problem 18CQ
Problem 19CQ
Problem 20CQ
Problem 21CQ
Problem 22CQ
Problem 23CQ
Problem 24CQ
Problem 25CQ
Problem 26CQ
Problem 27CQ
Problem 1PE
Problem 2PE
Problem 3PE
Problem 4PE
Problem 5PE
Problem 6PE
Problem 7PE
Problem 8PE
Problem 9PE
Problem 10PE
Problem 11PE
Problem 12PE
Problem 13PE
Problem 14PE
Problem 15PE
Problem 16PE
Problem 17PE
Problem 18PE
Problem 19PE
Problem 20PE
Problem 21PE
Problem 22PE
Problem 23PE
Problem 24PE
Problem 25PE
Problem 26PE
Problem 27PE
Problem 28PE
Problem 29PE
Problem 30PE
Problem 31PE
Problem 32PE
Problem 33PE
Problem 34PE
Problem 35PE
Problem 36PE
Problem 37PE
Problem 38PE
Problem 39PE
Problem 40PE
Problem 41PE
Problem 42PE
Problem 43PE
Problem 44PE
Problem 45PE
Problem 46PE
Problem 47PE
Problem 48PE
Problem 49PE
Problem 50PE
Problem 51PE
Problem 52PE
Problem 53PE
Problem 54PE
Problem 55PE
Problem 56PE
Problem 57PE
Problem 58PE
Problem 59PE
Problem 60PE
Problem 61PE
Problem 62PE
Problem 63PE
Problem 64PE
Problem 65PE
Problem 66PE
Problem 67PE
Problem 68PE
Problem 69PE
Problem 70PE
Problem 71PE
Problem 72PE
Problem 73PE
Problem 74PE
Problem 75PE
Problem 76PE
Problem 77PE
Problem 78PE
Problem 79PE
Problem 80PE
Problem 81PE
Problem 82PE
Problem 83PE
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College Physics

1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000
Textbook Problem

Data from the appendices and the periodic table may be needed for these problems.

What fraction of the 40K mat was on Earth when it formed 4.5 × 10 9 years ago is left today?

To determine

To fraction of K40 left on Earth after 4.5×109 years.

Explanation

Given info:

  • The time elapsed is 4.5×109years.
  • The half-life of K40 is 1.28×109 y .

Formula used:

Formula to find the left fraction is,

  NiN= e0.693 t 1 2×t

Here, Ni is the initial number of K40 , N is the present number of atom, t12 is the half-life of K40 and t is the time elapsed

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Ch-31 P-1CQCh-31 P-2CQCh-31 P-3CQCh-31 P-4CQCh-31 P-5CQCh-31 P-6CQCh-31 P-7CQCh-31 P-8CQCh-31 P-9CQCh-31 P-10CQ
Ch-31 P-11CQCh-31 P-12CQCh-31 P-13CQCh-31 P-14CQCh-31 P-15CQCh-31 P-16CQCh-31 P-17CQCh-31 P-18CQCh-31 P-19CQCh-31 P-20CQCh-31 P-21CQCh-31 P-22CQCh-31 P-23CQCh-31 P-24CQCh-31 P-25CQCh-31 P-26CQCh-31 P-27CQCh-31 P-1PECh-31 P-2PECh-31 P-3PECh-31 P-4PECh-31 P-5PECh-31 P-6PECh-31 P-7PECh-31 P-8PECh-31 P-9PECh-31 P-10PECh-31 P-11PECh-31 P-12PECh-31 P-13PECh-31 P-14PECh-31 P-15PECh-31 P-16PECh-31 P-17PECh-31 P-18PECh-31 P-19PECh-31 P-20PECh-31 P-21PECh-31 P-22PECh-31 P-23PECh-31 P-24PECh-31 P-25PECh-31 P-26PECh-31 P-27PECh-31 P-28PECh-31 P-29PECh-31 P-30PECh-31 P-31PECh-31 P-32PECh-31 P-33PECh-31 P-34PECh-31 P-35PECh-31 P-36PECh-31 P-37PECh-31 P-38PECh-31 P-39PECh-31 P-40PECh-31 P-41PECh-31 P-42PECh-31 P-43PECh-31 P-44PECh-31 P-45PECh-31 P-46PECh-31 P-47PECh-31 P-48PECh-31 P-49PECh-31 P-50PECh-31 P-51PECh-31 P-52PECh-31 P-53PECh-31 P-54PECh-31 P-55PECh-31 P-56PECh-31 P-57PECh-31 P-58PECh-31 P-59PECh-31 P-60PECh-31 P-61PECh-31 P-62PECh-31 P-63PECh-31 P-64PECh-31 P-65PECh-31 P-66PECh-31 P-67PECh-31 P-68PECh-31 P-69PECh-31 P-70PECh-31 P-71PECh-31 P-72PECh-31 P-73PECh-31 P-74PECh-31 P-75PECh-31 P-76PECh-31 P-77PECh-31 P-78PECh-31 P-79PECh-31 P-80PECh-31 P-81PECh-31 P-82PECh-31 P-83PE

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