# Fusion power

Suppose your car uses 430 gallons of gasoline per year, with each gallon producing $1.2*10^8 J$ of energy. If you had a fusion-powered car using the reaction 2H + 3H = 1n + 4He what mass of 2H would be required instead of that 430 gallons of gasoline? ((B)) What mass of 3H would be required instead of that 430 gallons of gasoline?

Energy of gasoline 430 gallons of is

$E1 = 430*1.2*10^8 J = 5.16*10^{10} J$

$(2,1)H + (3,1) H = (1,0)n + (4,2)He$

Atomic masses of atoms are

$M(2,1)H = 2.0141018 amu$

$M(3,1)H = 3.0160493 amu$

$M(1,0)n =1.0086649 amu$

$M(4,2)He = 4.002603 amu$

mass defect in above reaction is

$\Delta(m) = -4.002603 -1.0086649 +3.0160492+ 2,0141018 =0.018831 amu$

$E = m*c^2 = 0.01883*1.66*10^{-27}*9*10^16 =2.82*10^-{12} J$

For a total energy of E1 the number of nucleus 3H required are

total number of

$N = E1/E =1.83*10^{22}$ 3Hydrogen nucleus

Number of moles of 3H requires is (Na is Avogadro number)

$n = N/Na = 1.83*10^22/6.023*10^23 =0.0304 moles$

Mass of one 3H mole is M(3H) = 3 g

Thus total mass of 3H required is

$m = M*n = 3*0.0304 =0.091 grams$