Ford Nucleon -- Fission Powered Car Calculation

[ninne]

Homework Statement

1. In 1958, the Ford Nucleon was launched, a car powered by nuclear energy. It had a small reactor onboard. Assume the car had a mass of 1100 kg and that the reaction used to release energy in the reactor was the following:

U-235 → U-236 → Sr-95 + Xe-139 + 2n

How many grams of uranium-235 are needed for the car to accelerate from 0 km/h to 25 km/h? Assume the nuclear reactor has an efficiency of 30%.

Relevant Equations

E = mc^2

Jr

 

[renormalize]

Please show your attempt to solve the problem.

 

[ninne]

We are asked to calculate how many grams of uranium-235 are required to accelerate a 1100 kg car from 0 to 25 km/h, assuming:

• The nuclear reactor uses the fission of uranium-235:
  U-235 + n → U-236* → Sr-95 + Xe-139 + 2n
• The efficiency of the nuclear engine is 30%
• The energy released per fission is:
  E_fission = 3.885 × 10^-11J

v = 25 km/h ÷ 3.6 = 6.944 m/s


2. Calculate the car’s kinetic energy


E_kin = (1/2) * m * v²


E_kin = 0.5 * 1100 kg * (6.944 m/s)² ≈ 26,459 J


3. Account for reactor efficiency


Only 30% of the nuclear energy becomes kinetic energy:


E_needed = E_kin ÷ 0.30 = 26,459 J ÷ 0.30 ≈ 88,197 J





4. Calculate number of fissions required


Each fission releases 3.885 × 10⁻¹¹ J:


N = 88,197 J ÷ 3.885 × 10⁻¹¹ J ≈ 2.271 × 10¹⁵ fissions


5. Convert number of fissions to grams of U-235


First, convert fissions to moles using Avogadro’s number:


n = 2.271 × 10¹⁵ ÷ 6.022 × 10²³ ≈ 3.772 × 10⁻⁹ mol


Then, convert moles to grams using molar mass of U-235 (235 g/mol):


m = 3.772 × 10⁻⁹ mol × 235 g/mol ≈ 8.886 × 10⁻⁷ g

 

[renormalize]

m = 3.772 × 10⁻⁹ mol × 235 g/mol ≈ 8.886 × 10⁻⁷ g

Looks good to me.