PHYS 106 Final Method Review P10 Werk 4. The primary neutron-producing isoeope o

Question

PHYS 106 Final Method Review P10 Werk 4. The primary neutron-producing isoeope of uranium loaded in the Little Boy bomb dropped on Hiroshima was electrons as hydrogen. How many a. Neutral uranium carries 92 times as many neutrons are in unani b The mass of the uranium-235 nucleus is 235.0439299 u, the rest massofa is 1.007 276 5 w and the rest mass of a neutron is 1.0086649 unlu 1,660 5390 x10 kg and c 2997 9246 x 10 m/s, what is the binding energy of a single uranium-235 nucleus in Mev? r 2.30 947 assembled and c. The Little Boy bomb was loaded with 64 kg of uranium-235, presumably weighed on June 15 and delivered on August 6, 1945. i. How many uranium-235 nuclei were in this 64 kg? ii. Assuming the same time of day for assembling and dropping, how many z sr seconds passed from June 15 to August 6? iii. If the half-life of uranium-235 is7.038x10" y, what is the halflife in iv. what percentage of uranium-235 remained at the time the bomb was dropped compared to when it was assembled? d. The fission ofone nucleus of uranium-235 generates 3.24 x10 of energy which is bound up in the kinetic energy of the three neutrons that are released during fission. i. What is the energy in one of the released neutrons?

Explanation / Answer

4.

a. number of neutrons in U235 = 235 - 92 = 143

b. with 92 protons and 143 neutrons

mass = 92 x 1.0072765 + 143 x 1.0086649 = 236.90852 u

mass defect (dm) = 236.90852 - 235.0439299 = 1.8646

dm in kg = 1.8646/6.023 x 10^26 = 3.096 x 10^-27 kg

binding energy = (3.096 x 10^-27)(3 x 10^8)^2 = 2.79 x 10^-10 J

binding energy in MeV = 2.79 x 10^-10 J/1.602 x 10^-13 x 235

                                     = 7.401 MeV/nucleon

c. U-235 bomb

i. Number of nuclei = 64 x 6.023 x 10^23/235 = 1.64 x 10^23

ii. Time elapsed = 21 d = 21 x 24 x 60 x 60 = 1,81,4400 s

iii. half-life in seconds = 7.038 x 10^8 d x 24 x 60 x 60 = 6.10 x 10^13 s

iv. amount remaining [A] from initial [A]o after this time,

ln[A] = ln(64) - (ln(2)/7.038 x 10^8)(21)

[A] = 63.999999

percent U235 remained = 99.999998%

d.

i. energy of one neutron = 3.24 x 10^-11/3 = 1.08 x 10^-11 J

ii. rest mass of neutron = 1.675 x 10^-27 kg

iii. velocity = sq.rt.(2 x 1.08 x 10^-11/1.675 x 10^-27) = 1.13 x 10^8 m/s

e.

i. heat needed = 1100 x 440 x (1726 - 300) = 6.902 x 10^8 J

ii. Energy released from U235 = 2.79 x 10^-10 x 1.38 x 6.023 x 10^23/100 = 1.68 x 10^14 J/mol

iii. percentage of energy used to melt = 6.902 x 10^8 x 100/1.68 x 10^14 = 0.0004%

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