E4-1. Spontaneous & Induced Fission HL

Example 1: Fission Reaction Q-Value & Fragment Kinematics

Problem: A thermal neutron is absorbed by a U-235 nucleus, triggering fission: $${}^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \longrightarrow {}^{144}_{56}\text{Ba} + {}^{89}_{36}\text{Kr} + 3 {}^{1}_{0}\text{n}$$ Given the isotopic masses: U-235 = $235.0439\text{ u}$; Ba-144 = $143.9229\text{ u}$; Kr-89 = $88.9176\text{ u}$; Neutron = $1.0087\text{ u}$.
(a) Calculate the mass defect ($\Delta m$) and total energy released ($Q$-value) in MeV.
(b) By applying the conservation of momentum (assuming the original U-235 and incident neutron have negligible initial momentum compared to the explosive fragments), estimate the initial kinetic energy of the Barium-144 fragment. Assume the fragments share exactly $170\text{ MeV}$ of the total $Q$-value.

Solution:

Part (a): Mass Defect & Q-Value
Sum the reactant masses: $235.0439 + 1.0087 = 236.0526\text{ u}$
Sum the product masses: $143.9229 + 88.9176 + 3(1.0087) = 235.8666\text{ u}$
Mass defect: $\Delta m = 236.0526 - 235.8666 = \mathbf{0.1860\text{ u}}$
$Q = 0.1860\text{ u} \times 931.5\text{ MeV/u} \approx \mathbf{173.3\text{ MeV}}$

Part (b): Fragment Kinematics
To conserve zero initial momentum, the two heavy fragments must recoil in opposite directions with equal and opposite momenta ($p_{\text{Ba}} = p_{\text{Kr}}$). Because kinetic energy is $K = p^2/2m$, kinetic energy is inversely proportional to mass.
The fraction of kinetic energy taken by the heavier Barium fragment is: $$K_{\text{Ba}} = \left( \frac{m_{\text{Kr}}}{m_{\text{Ba}} + m_{\text{Kr}}} \right) K_{\text{total}}$$ $$K_{\text{Ba}} \approx \left( \frac{89}{144 + 89} \right) \times 170\text{ MeV} = \left(\frac{89}{233}\right) \times 170 \approx \mathbf{64.9\text{ MeV}}$$ Notice how the lighter Krypton fragment moves faster and takes the larger energy share ($\sim 105.1\text{ MeV}$).

Example 2: Liquid Drop Model & The Fission Coulomb Barrier

Problem: The "liquid drop model" of the nucleus suggests that upon absorbing a neutron, the U-236* compound nucleus elongates into a "dumbbell" shape before splitting. Assume it splits into two identical spherical Palladium fragments ($A \approx 118$, $Z = 46$).
Using the Fermi radius $R_0 = 1.20\text{ fm}$, estimate the electrostatic Coulomb repulsion potential energy (in MeV) between the two fragments the instant they separate (when their surfaces are just touching). This value estimates the kinetic energy barrier of the fission event.

Solution:

Step 1: Determine the physical radius of a single fragment.
$$R = R_0 A^{1/3} = (1.20 \times 10^{-15}\text{ m}) \times (118)^{1/3} \approx 1.20 \times 4.905 = 5.886 \times 10^{-15}\text{ m}$$

Step 2: Distance between charge centers.
When the two identical spheres are just touching, the distance $d$ between their centers is exactly $2R$. $$d = 2 \times 5.886\text{ fm} = 11.77\text{ fm} = 1.177 \times 10^{-14}\text{ m}$$

Step 3: Calculate the electrostatic potential energy ($E_p$).
$$E_p = \frac{k q_1 q_2}{d} = \frac{k (Z e)^2}{d} = \frac{(8.99 \times 10^9) \times (46 \times 1.60 \times 10^{-19})^2}{1.177 \times 10^{-14}}$$ $$E_p = \frac{(8.99 \times 10^9) \times (5.417 \times 10^{-35})}{1.177 \times 10^{-14}} = 4.138 \times 10^{-11}\text{ J}$$

Step 4: Convert to MeV.
$$E_p = \frac{4.138 \times 10^{-11}\text{ J}}{1.60 \times 10^{-13}\text{ J/MeV}} \approx \mathbf{258\text{ MeV}}$$ Context: This immense repulsive energy is what physically violently forces the fragments apart, converting internal electrical potential energy into the devastating kinetic energy of a nuclear explosion or power plant!

Example 3: Energy Yield from BE/A Data

Problem: Without using exact isotopic masses, estimate the energy released when a single nucleus of Uranium-235 ($A=235$) fissions into two roughly equal fragments of $A \approx 117$. Use the standard Binding Energy per Nucleon (BE/A) curve values: $BE/A$ for U-235 is $\sim 7.6\text{ MeV}$, and $BE/A$ for $A=117$ is $\sim 8.5\text{ MeV}$.

Solution:

Fission moves a nucleus up the BE/A curve to a more stable state. The energy released is simply the difference in total binding energy.
Total initial BE of U-235: $235 \times 7.6\text{ MeV} = 1786\text{ MeV}$
Total final BE of the fragments: $235 \times 8.5\text{ MeV} = 1997.5\text{ MeV}$
Energy Released = Final BE - Initial BE = $1997.5 - 1786 = \mathbf{211.5\text{ MeV}}$.
This is a remarkably fast, highly accurate method commonly tested on IB Paper 1 multiple-choice questions.