Ratio of Uranium in a Specific Sample

Assignment 3

Question 1.

Ratio of molecules in initial concentration

235U/238U in a particular rock sample was 2.0%

235U half-life = 7.04 x 108 238U half-life = 4.5 x 109

4.5 billion years = 4.5 x 109

number of half-lives occurred = (total time elapsed ÷ length of half-life)

(4.5 x 10^9)/( 7.04 x 10^8) = 6.42 half lives (4.5 x 10^9)/(4.5 x 10^9) = 1 half life

(starting amount) x (1/2)number of half-lives

After 6.42 half-lives the fraction of U235 = 2 / 26.42 = .0234

After 1 half-life the fraction of U238 = 100 / 21 = 50

0.0234:50 = .046% U235

Question 2.

A single U-235 atom releases approximately 200 MeV (million electron volt) http://science.howstuffworks.com/nuclear-power1.htm

235g of U-235 contains Avogadro's number of atoms and 1 kg contains-

1000g X (6.02 x 1023 2 2 235g =2.563x 1024 Atoms

Energy released = # of atoms X energy released per fission

2.563 x 1024 X 200 MeV = 5.126 X 1026 MeV

1 MeV = 106 x 1.6 x 10-19 = 1.6 x 10-13 J http://www.a-levelphysicstutor.com/nucphys-fission.php

The energy released is 1.6 x 10-13 J x 5.126 X 1026 MeV = 8.2X 1013 J = 8.2X 1010 KJ

1 CH4 = (802 Kj )/(mol CH4) X (mol CH4)/16g = 50.12 Kj / CH4

1 kg = 1000g = 50.1 Kj x 1000g = 5.01 x 104 Kj / Kg

Ratio ( 8.20X 10^10 Kj)/(5.01 x 10^4 Kj ) = 1.6 X 106 U-235 : 1 CH4

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