Fyneman exercise II ch1-2

Q.

Make a rough estimate of the work that must be done against electric force to assemble a uranium nucleus from two equal halves. What about assemble two deuterons to make a helium nucleus? Also express both answers in kilowatt hours per kg.

Trial.

Equation of nuclear radius from wiki(Atomic nucleus)
$R=R_0A^{1 \over 3}\tag{1}$

, where $A$ is atomic mass number and values for calculation is here.

Uranium

Atomic number of uranium is 92, which means nuleus has 92 protons.
So, one half of it has 46 protons.

From eq.(1),
$R_U = 1.25 \times 10^{-15}\ m \times 238^{1 \over 3} = 7.75\times 10^{-15}\ m$

Suppose we move two halves of uranium with 46 protons from $\infty$ to $R_U$.

$W_{done}=-{1 \over 4\pi\epsilon_0}\int_{\infty}^{R_U}{(46q)^2 \over r^2}dr={1 \over 4\pi\epsilon_0}\left[{(46q)^2 \over r}\right]^{R_U}_{\infty}={1 \over 4\pi\epsilon_0}{(46q)^2 \over R_U}$

$W_{done}={1 \over 4\times 3.14 \times 8.85\times10^{-12}}{(46\times 1.06)^2 \times 10^{-38}\over 7.75\times 10^{-15}}\approx5.18\times 10^{-8} \ J$

We need about $5.18 \times 10^{-8}\ J$.

$Atomic\ mass\ of\ uranium = 238\ u = 238 \times 1.66\times 10^{-27}\ kg$

$Number\ of\ uranium\ atoms\ in\ 1kg = {1 \over 238 \times 1.66\times 10^{-27}}$

$E_{/ kg}={5.18\times 10^{-8}\ J \over 3.95 \times 10^{-25}}=1.31 \times 10^{17}\ J$

Kilowatt-hour=$3.6\times 10^6 \ J$

So, energy in killowatt-hours per kg

${1.31\times 10^{17}\over 3.6\times10^6}=3.64\times10^{10}$

Helium

Atomic number of uranium is 2, which means nuleus has 2 protons.
So, one half of it has 1 protons.

From eq.(1),
$R_D = 1.25 \times 10^{-15}\ m \times 4^{1 \over 3} = 1.98\times 10^{-15}\ m$

$W_{done}={1 \over 4\times 3.14 \times 8.85\times10^{-12}}{( 1.06)^2 \times 10^{-38}\over 1.98\times 10^{-15}}\approx5.11\times 10^{-14} \ J$

We need about $5.11 \times 10^{-14}\ J$.

$Atomic\ mass\ of\ uranium = 4\ u = 4 \times 1.66\times 10^{-27}\ kg$

$Number\ of\ helium\ atoms\ in\ 1kg = {1 \over 4\times 1.66\times 10^{-27}}$

$E_{/ kg}={5.11\times 10^{-14}\ J \over 1.31 \times 10^{-27}}=3.90 \times 10^{13}\ J$

Kilowatt-hour=$3.6\times 10^6 \ J$

So, energy in killowatt-hours per kg

${3.90\times 10^{13}\over 3.6\times10^6}=1.08\times10^{7}$