Question

Using fission, what mass m of uranium-235 would be necessary to supply all of the energy that the united states uses in a year, roughly 1.0×1019j

Answer 1

The energy released by 1 atom of Uranium is Q=185.*10^6-25.
The energy needed for U.S. that I will name Q2, in eV is: 1*10^19J* (1eV/(1.602*10^-19))=6.2422*10^37eV

Q=mc^2
Q2=mc^2

m is the mass of Uranium in kg that we calculated at 3.90*10^-25. We need to fund m2 because that is the unknown mass in the equation for Q2 to give us the energy for the U.S. and remember, Q is the energy that 1 atom of Uranium gives out which is why we took a ratio.


m2=Q2/Q*m=1.3*10^5 kg (final answer)

Answer 2

1.22205 × 10⁵kg of Uranium would be required

Step-by-step explanation:

It is important to know that there are various ways in which Uranium can undergo fission to produce energy. Depending on the type of fission it undergoes, the energy released might be diffferent.

A typical fission reaction of uranium –235 is:

²³⁵₉₂U + ¹₀n → ¹³⁹₅₇La + ⁹⁵₄₂Mo + 2 ¹₀n + 7 ⁰₋₁e

In energy fission, energy released is calculated from the formular;

Energy, E = m × c²

Where m = mass loss during fission

c = speed of light

To obtain mass loss, we compare the the mass in the reactant side with that of the product side.

Total mass of reactants = Mass of U + Mass of 1 neutron

= 235.124 + 1.009 = 236.133 u.....(1)

Total mass of products = Mass of La + Mass of Mo + Mass of 2 neutrons

= 138.955 + 94.946 + 2 × 1.009

= 138.955 + 94.946 + 2.018 = 235.919 u....(2)

Loss of mass during fission = Mass of reactants – Mass of products

= 0.214 u . . . . . . . (3)

This mass of 0.214 atomic mass units is now to be converted into kilograms.

1 u = 1.66 × 10⁻²⁷ kg

0.214 u = x

Upon cross multiplication;

x = 1.66 × 10⁻²⁷ × 0.214 kg

x = 0.355 × 10⁻²⁷ kg

Thus, the mass lost (m) from 1 nucleus (or 1 atom) of uranium –235 is 0.355 × 10⁻²⁷ kg.

Energy, E = m × c² = [0.355 × 10⁻²⁷] × [(3 × 10⁸)²] joules = 3.195 × 10⁻¹¹ joules.

This much energy is released by the fission of one uranium –235 nucleus.

We will now calculate the energy released by the fission of 1 mole of uranium –235 nuclei or 6.023 × 10²³ uranium –235 nuclei.

Energy released by the fission of 1 U – 235 nucleus = 3.195 × 10⁻¹¹ J

So, Energy released by the fission of 6.023 × 10²³ U –235 nuclei = 3.195 × 10⁻¹¹ × 6.023 × 10²³ J = 19.24 × 10¹² J.

This much energy is released by the fission of 1 mole of uranium –235 which is equal to gram atomic mass of uranium –235 or 235.124 grams of uranium –235.

But we have to calculate the mass of uranium necessary to supply all of the energy that the united states uses in a year, roughly 1.0×10¹⁹ j

Now, Energy released from 235.124 g of uranium = 19.24 × 10¹² J.

x mass of uranium = 1.0×10¹⁹ J.

Upon cross multiplication we have;

x = (235.124 * 1.0×10¹⁹ ) / 19.24 × 10¹²

x = 122, 205, 821.21 g = 122,205.82 kg

Thus, mass m of uranium-235 necessary to supply all of the energy that the united states uses in a year is 1.22205 × 10⁵kg