Astrophysics for IB Physics HL

Reading cue. Count what crosses the drawing: one neutron in, two fragments out, nothing else. The free neutrons that carry the chain on are not drawn, so the nucleon numbers as pictured do not balance.

1. aState State the form in which almost all of the 200 MeV appears immediately after the split, and state what the two arrows drawn on the fragments show about the momentum of the system.

   recall2 marks

   Check answer · 2 marks

   1. Almost all appears as kinetic energy of the two fragments, driven apart by the electrostatic repulsion between their positive charges
   2. The arrows point in opposite directions, so the fragments carry equal and opposite momenta and the total momentum stays essentially that of the slow incoming neutron, close to zero

2. bDetermine Determine the energy released in the single event drawn, in joules, and hence determine the number of such events needed each second to sustain a thermal output of 3.2 GW.

   routine4 marks

   Check answer · 4 marks

   1. 200 MeV = 200 × 10⁶ × 1.60 × 10⁻¹⁹ J
   2. = 3.2 × 10⁻¹¹ J
   3. Number per second = 3.2 × 10⁹/(3.2 × 10⁻¹¹)
   4. = 1.0 × 10²⁰ events per second

3. cDetermine The two fragments drawn have nucleon numbers 141 and 92, and about 170 MeV of the energy released appears as their kinetic energy. Determine the kinetic energy carried by each fragment.

   demanding4 marks

   Check answer · 4 marks

   1. Momentum conservation gives the two fragments equal and opposite momenta
   2. Since kinetic energy equals p²/2m, the energy is shared in inverse proportion to mass, so the shares are in the ratio 141 : 92 in favour of the lighter fragment
   3. Lighter fragment, nucleon number 92: 170 × 141/233 = 103 MeV
   4. Heavier fragment, nucleon number 141: 170 × 92/233 = 67 MeV

4. dDiscuss Nothing is drawn leaving the reaction except the two fragments. Discuss what else must leave the nucleus, and what decides whether the single event drawn grows into a self-sustaining chain.

   top of the paper4 marks

   Check answer · 4 marks

   1. Nucleon numbers must balance: 235 + 1 = 236, while 141 + 92 = 233, so three free neutrons must also be released
   2. A chain is sustained only if, on average, exactly one neutron from each fission goes on to cause a further fission
   3. The remaining neutrons are lost by escaping through the surface or by being absorbed without causing fission, so the mass and shape of the sample decide the outcome
   4. The released neutrons are fast and are captured by ²³⁵U far less readily until repeated collisions in a moderator have slowed them