# Fission

## Practice

### practice problem 1

#### solution

Answer it.

### practice problem 2

#### solution

Answer it.

### practice problem 3

#### solution

Answer it.

### practice problem 4

#### solution

Start with the equation for electrostatic potential energy (*U _{e}*) between two point charges.

U = _{e} |
kq_{1}q_{2} |

r |

This will be computed over and over again for every pair of protons in the nucleus. All protons are identical, so…

q_{1} = q_{2} = e |
& | q_{1}q_{2} = e^{2} |

Thus…

U = ∑ _{parent} |
ke^{2} |

r |

Dividing a ball of nuclear matter in half results in two spheres with half as many charges and half as much volume. Volume is proportional to the cube of radius, so the new spheres will have a diameter that's smaller by the cube root of a half.

V ∝ r^{3} |
⇒ | V | ∝ | ⎛ ⎜ ⎝ |
r |
⎞^{3}⎟ ⎠ |

2 | ∛2 |

Applying these changes to the electrostatic potential energy for one daughter nucleus gives…

U = _{daughter} |
∑ | k(e/2)^{2} |
= | ∛2 | ∑ | ke^{2} |

r/∛2 |
4 | r |
||||

U = _{daughter} |
31.5% U_{parent} |

Thus, each daughter nucleus has 31.5% of the electrostatic energy of the parent nucleus. But…

31.5% + 31.5% = 63% ≠ 100%

The remaining 37% is liberated to do work — like boiling water to make steam to generate electricity or throwing out a blast wave to further a military agenda.