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Physics Energy Calculations Guide

This document contains 11 physics questions and solutions related to nuclear power and uranium fission. Question 3 summarizes that uranium fission produces almost 10,000 times more energy than combustion of carbon, showing uranium has a much greater energy density and can generate energy for longer. Question 6 calculates that the centrifuges used to separate uranium isotopes must spin with a centripetal force over 10 million times the weight of a uranium molecule, equivalent to 16.2 giganewtons on a 50kg student.

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Bosco Merino Echevarria
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110 views1 page

Physics Energy Calculations Guide

This document contains 11 physics questions and solutions related to nuclear power and uranium fission. Question 3 summarizes that uranium fission produces almost 10,000 times more energy than combustion of carbon, showing uranium has a much greater energy density and can generate energy for longer. Question 6 calculates that the centrifuges used to separate uranium isotopes must spin with a centripetal force over 10 million times the weight of a uranium molecule, equivalent to 16.2 giganewtons on a 50kg student.

Uploaded by

Bosco Merino Echevarria
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOC, PDF, TXT or read online on Scribd
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GEP 17

Solutions Nuclear Power

Q1. No atoms U-235 = 0.007 x 1000 x 6.02 x 1023 / 235 = 1.79 x 1022
energy available = 1.79 x 1022 x 200 MeV = 1.79 x 1022 x 200 x 1.60 x 10-13 = 5.73 x 1011 J

Q2. No moles C-12 = 1000 / 12 = 83.3


energy available = 83.3 x 0.39 MJ = 79.9 MJ

Fission energy / combustion energy = 5.73 x 1011 / 79.9 x 106 = 7170

Q3. The energy available from uranium fission is almost 10,000 times greater than from the
combustion of a comparable amount of carbon. This suggests that uranium has a much
greater energy density and therefore can generate energy for a much longer time.

Q4. Relative atomic mass UF6 = 238 + 6 x 19 = 352


mass of one molecule = 352 x 1.66 x 10-27 kg = 5.84 x 10-25 Kg.
(The answer depends on the value taken for 1 amu)

Q5. (i) no revs per second = 70000 / 60 = 1170 s-1


(ii) Period; T = 1/1170 = 8.55 x 10-4 s
(iii) angular velocity;
2 2
  4
 14700 s 1
T 8.55  10

Q6. (i) F  mr 2  5.84 10 25 1.5  147002  1.89 10 16 N
(ii) F / W = 1.89 x 10-16 / (5.84 x 10-25 x 9.81) = 3.30 x 107
(iii) This ratio is a huge number, the centripetal force being more than 10 million times the
weight of the molecule illustrating the incredibly high speeds at which the centrifuges must
spin in order to separate out the Uranium isotopes. An equivalent force on a 50Kg student
= 50 x 9.81 x 3.30 x 107 = 1.62 x 1010 N (16.2 GN!)

Q7. Probability = a / A

Q8. Probability = b/ A

Q9. (i) Ratio = a / b (ii) This is the ratio of probabilities of a collision with “a” compared with ”b”

1000
Q10.   200
5

100
Q11.   0.01
10,000

Side 1 of 1

Sha Tin College Physics  OUP IB Physics Course companion, 2007, Kirk and Hodgson. These pages have been
adapted from the original 1

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