Question

Challenge problem:Critical mass is the smallest amount of radioactive material necessary to sustain a nuclear chain reaction. Uranium-235, the isotope of uranium used in the "little boy" bomb dropped on Hiroshima during Wordl War II,has a critical mass of 52 kg. How many long tons of yellowcake (64% uranium by mass) need to be processed to produce a critical mass of U-235? The percent natural abundance of U-235 is 0.72% and its nuclide mass is 235.0439 amu.

1 long ton = 2240 lb., 1 kg = 2.20 lb.

Answer 1

Let the mass of yellowcake required to produce 52kg (critical mass) of U-235 be 'x' kg

The mass of Uranium in the above mass of yellowcake = (64% of x) kg = '0.64x' kg

Now, the number of atoms of U-235 needs to be found in the above mass using the percent abundance of U-235 which is 0.72%

The number of Uranium g-atoms in '0.64x' kg of Uranium = 0.64x kg (1000g/1kg) (1g-atom/238.02891g) = 2.68875x

{Using the average atomic mass of U (238.02891u) and the fact that 1 g-atom of U (=238.02891g)}

The number of g-atoms of U-235 in the above number = (0.72% of 2.68875x) = 0.019359x

1g-atom of U-235 has a mass of 235.0439g (from nuclide mass)

The mass of 0.019359x g-atoms of U-235 = (0.019359x g-atoms of U-235) (235.0439g of U-235/1g-atom of U-235)

=> The mass of 0.019359x g-atoms of U-235 = 4.5502x g of U-235 = 4.5502x g (1kg/1000g) of U-235

=>The mass of 0.019359x g-atoms of U-235 = 0.0045502x kg of U-235

The above value must be equal to 52 kg (according to the condition of the question)

0.0045502x = 52

=> x = 11428.06910

i.e. 11428.06910 kg of yellowcake needs to be processed to get 52kg (critical mass) of U-235.

Converting this value to long tons:

11428.06910 kg = 11428.06910 kg (2.20 lb/1kg) (1 long ton/2240 lb) = 11.2240 long ton

The mass of yellowcake that needs to be processed = 11.2240 long ton