**(a)**

From the law of conservation of energy:

$\overline{){{\mathbf{K}}}_{\mathbf{1}\mathbf{H}}{\mathbf{+}}{{\mathbf{K}}}_{\mathbf{2}\mathbf{H}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{-}}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi}{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{2}\mathbf{r}}}$

The deuterium nucleus starts out with a kinetic energy of 1.09x10^{-13} joules, and the proton starts out with a kinetic energy of 2.19x10^{-13} joules. The radius of a proton is 0.9x10^{-15} m; assume that if the particles touch, the distance between their centers will be twice that.

a) What will be the total kinetic energy of both particles an instant before they touch?

K_{1H} + K_{2H} =

b) What is the kinetic energy of the reaction products (helium nucleus plus photon)? K_{(He)} + K_{(gamma)} =

c) Fusion as energy source: Kinetic energy can be used to drive motors and do other useful things. If a mole of hydrogen and a mole of deuterium underwent this fusion reaction, how much kinetic energy would be generated? joules (For comparison, around 1x10^{6} joules are obtained from burning a mole of gasoline.)

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