Free Response Practice Question on Nuclear Physics for AP Physics B

Occasionally you may get a free response question involving nuclear physics. Here is a question meant for exercise. This question carries 15 points and you can take 17 minutes for answering it. Try to answer the question yourself before going through the solution given below the question.

One proposal for the production of power is the fusion of deuterium to produce tritium in accordance with the nuclear reaction

₁D² + ₁D² → ₁T³ + ₁H¹ + Q

where the symbols D and T represent respectively deuterium and tritium. Mass of ₁D² = 2.0141 u; Mass of ₁T³ = 3.0161 u; Mass of ₁H¹ = 1.0078 u and 1 u = 1.66×10^–27 kg = 931 MeV/c²

Now answer the following:

(a) Calculate the mass defect involved in a single fusion reaction.

(b) Determine the energy Q released in a single fusion reaction.

(c) Calculate the number of fusions required per second to produce a power of 2000 mega watt.

(d) If the efficiency of the process is assumed to be 60%, determine the mass of deuterium required per day.

(a) Mass defect per fusion ∆m = (2×Mass of ₁D²) – (Mass of ₁T³ + Mass of ₁H¹)

= (2×2.0141 u) – (3.0161 u + 1.0078) = 0.0043 u

(b) Energy released in a single fusion reaction = 0.0043×931 MeV = 4.0033 MeV

(c) Since 1 MeV = 1.6×10^–13 J, the number (n) of fusions required per second to produce a power of 2000 mega watt (which is 2000×10⁶ joule per second) is given by

n = (2000×10⁶)/( 4.0033×1.6×10^–13) = 3.1224×10²¹

(d) Number of fusions required per day = 3.1224×10²¹×(24×60×60) = 2.6977×10²⁶.

Since two deuterons are required per fusion, the number of deuterons required per day (at 100% efficiency) is 2×2.6977×10²⁶ = 5.3954×10²⁶.

Since the efficiency of the process is 60% only, the required number of deuterons is 5.3954×10²⁶ ×(100/60) = 8.9923×10²⁶.

Since 1 u = 1.66×10^–27 kg, the mass of the above number of deuterium atoms = 8.9923×10²⁶×2.0141×1.66×10^–27 kg = 3.0065 kg.

Now, suppose the part (d) in the above question is replaced by the following:

(d) If the radius of the deuteron is assumed to be nearly 1.5×10^–15 m, calculate (i) the kinetic energy required by a deuteron to overcome the coulomb repulsion and (ii) the temperature to which the deuterium gas is to be heated to initiate the fusion reaction. Boltzmann’s constant, k = 1.38×10^–23 J/K

When the deuterons are in contact, the separation (r) between their centres is 2×1.5×10^–15 m (= 3×10^–15 m) and the electrostatic potential energy is (1/4πε₀)q²/r where ε₀ is the permittivity of free space and q is the charge on the deuteron (which is equal to 1.6×10^–19 coulomb). The kinetic energy required to overcome the coulomb repulsion must be at least equal to the above potential energy.

Therefore, kinetic energy = (1/4πε₀)q²/r = 9×10⁹×(1.6×10^–19)² /3×10^–15 = 7.68×10^–14 J. Since the deuteron has 3 degrees of freedom its kinetic energy is (3/2)kT where k is Boltzmann’s constant and T is the absolute (Kelvin) temperature.

Therefore (3/2)kT = 7.68×10^–14 so that T = 2×7.68×10^–14/(3×1.38×10^–23) = 3.7×10⁹ K.