Nuclear homework presentation by Linarez

Problem 1 Molecular weight of U-235:  253.04 [𝑔 𝑚𝑜𝑙]

Natural uranium contains: 72%     U-235                    99.28% U-238

Amount of uranium contained in 1 [kg] natural uranium:

1 [kg] uranium = .72 [𝑔] 100 [𝑔] 𝑥 1000 [𝑔]=7.2 [𝑔] U-235

The number of moles U-235: 

𝑔𝑟𝑎𝑚𝑠 𝑈𝑟𝑎𝑛𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑈−235 = 7.2 [𝑔] 235.04 [𝑔 𝑚𝑜𝑙] = 0.03 [mol U-235]

The number of atoms U-235:  number of mol U-235 (Avagdros number) 

 = (0.03 [mol U-235]) x (6.022 x 1023 [𝑎𝑡𝑜𝑚 𝑚𝑜𝑙]) = 1.80 x 1022 [atoms U-235]

Total energy released in fissioning 1 [kg] of Uranium:

1 [atom] 𝑦𝑖𝑒𝑙𝑑𝑠 →     198 [MeV]

 (1.80 x 1022 [atoms U-235] ) x (198 [MeV]) = 3.56 x 1024 [MeV]

 (3.56 x 1024 [MeV]) x (1.6 x 10−13 [𝐽 𝑀𝑒𝑉]) = 5.6 x 1011 [J]

1 [g] u-235

𝑦𝑖𝑒𝑙𝑑𝑠 →    8.28 x 1010 [J]

1[kg] u-235

𝑦𝑖𝑒𝑙𝑑𝑠 →    8.28 x 1013 [J]

Problem 4  𝑈146 92 238 𝛽 𝑑𝑒𝑐𝑎𝑦 →      𝐶𝑠85 55 140 + 𝑅𝑏55 37 92 + (𝑋𝐶 𝐵 𝐴 )

  Uranium (U) Cesium (Cs) Rubidium (Rb) Cs + Rb 𝑋𝐶 𝐵 𝐴

Atomic Mass Number

238 140 92 232 6

Number of protons 

92 55 37 92 0

Number of neutrons

146 85 55 140 6  (𝑋𝐶 𝐵 𝐴 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 𝑙𝑎𝑤 →             6𝑛1 0 1 )

The Energy Produced By 1 [Kg] of burning coal:  1 [lb] coal 𝑦𝑖𝑒𝑙𝑑𝑠 →    1.3 x 104 [Btu]  ( 1.3 x 104 [𝐵𝑡𝑢 𝑙𝑏])𝑥 ( 1 [𝑙𝑏] 0.453 [𝑘𝑔])=28.6 𝑥 103[𝐵𝑡𝑢 𝑘𝑔]  1 [kg]of burning coal 𝑦𝑖𝑒𝑙𝑑𝑠 →     2.9 𝑥 104[𝐵𝑡𝑢 𝑘𝑔]  2 .9 𝑥 104[𝐵𝑡𝑢 𝑘𝑔] 𝑥 (1055 [𝐽] 1 [𝐵𝑡𝑢])=3.1 𝑥 107[𝐽 𝑘𝑔]

According to the energy resources report of 2011:   U−235 → 2.074 x 104 [𝑡𝑜𝑛𝑠]  coal → 1.9233 x 1010 [𝑡𝑜𝑛𝑠] Solution:

For Uranium  2.074 x 104 [𝑡𝑜𝑛𝑠] 𝑥 (106[𝑔] 1 [𝑡𝑜𝑛]) 𝑥 (8.28 𝑥 1010[𝐽] 1 [𝑔] )=1.71 𝑥 1022 [𝐽]

For Coal  1.9223 x 1010 [𝑡𝑜𝑛𝑠] 𝑥 (106[𝑔] 1 [𝑡𝑜𝑛]) 𝑥 (8.28 𝑥 1010[𝐽] 1 [𝑔] )=5.4 𝑥 1026 [𝐽]

By neutron capture, U-238 results in the formation of the Pu-239 isotope. Pu-239 is another fissionable nuclear fuel.

𝑈−238𝑛 𝑐𝑎𝑝𝑡𝑢𝑟𝑒 →        𝑈−239  𝑈−239 𝛽 𝑑𝑒𝑐𝑎𝑦 (𝐻𝑎𝑙𝑓 𝐿𝑖𝑓𝑒= 24 [𝑚𝑖𝑛] →                       𝑁𝑝−239  𝑁𝑝−239𝛽 𝑑𝑒𝑐𝑎𝑦 (𝐻𝑎𝑙𝑓 𝐿𝑖𝑓𝑒= 2.3 [𝑑𝑎𝑦𝑠] →                       𝑃𝑢−239 𝑃𝑢−239  →𝐻𝑎𝑙𝑓 𝐿𝑖𝑓𝑒= 24000 [𝑦𝑟] 1 3 𝑜𝑓 𝑡𝑜𝑡𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑒𝑛𝑒𝑟𝑔𝑦  60% 𝑜𝑓 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑖𝑛𝑔𝑠 𝑎𝑟𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑃𝑢−239 𝑖𝑠𝑜𝑡𝑜𝑝𝑒

• Pu-239 consists of 90-95% weapon-based plutonium • To make weapons, you must use a fissile isotope • Plutonium is one of the most fissionable elements 

Number of internal reactor events = (number of events) x (number of years) = probability

(10−6[𝑅𝑌−1]) 𝑥 (𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠)= 1 50 ( 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠)=3333 [𝑦𝑟𝑠]