I want to work out the accelerating voltage for the B11 proton reaction.

From this excellent discussion of

fusion reactions [pdf] we find that the optimum voltage for the p-B11 reaction when accelerating protons into B11 is 550 KV. Representing a proton energy of 550 Kev. However because we are accelerating protons and B11 from the same virtual grid voltage some adjustments must be made.

First off the reaction runs based on velocity not energy so we have to adjust the accelerating voltage to account for that.

The energy of a non-relativistic particle = 1/2 mV

^{2} However to keep from cluttering up the calculation we will throw out the 1/2 factor since we are not concerned with actual energy but just input voltage and for that the 1/2 cancels out. Masses will be given in Atomic Mass Units where the proton or neutron = 1. Also since we are not concerned with actual values but values relative to 550 Kev, energy will be given as Kev.

So let us start with proton velocity.

mV

^{2}=550Kev

Since the mass of a proton = 1 that simplifies to the required velocity = √550,000 = 741.6198 in arbitrary units.

For a given accelerating voltage the energy of the B11 will be 5X the energy of the proton because B11 has 5X the charge of a proton.

E

_{p}=V

_{p}^{2}E

_{B11}=11*V

_{B}^{2}V

_{p} + V

_{B} = √550,000

Since the energy of the B11 is 5X the energy of the p then

5*V

_{p}^{2} = 11*V

_{B}^{2}V

_{p}^{2} = (11/5)*V

_{B}^{2}V

_{p} = (√(11/5))*V

_{B}V

_{p} + V

_{B} = √550,000

substituting:

(√(11/5))*V

_{B} + V

_{B} = √550,000

calculating the result

V

_{B} = 298.650

V

_{p} = 442.970

The energy of the proton will be:

442.970

^{2} = 196,222.175 or an accelerating voltage of about 200KV

The energy of the boron will be:

11 * 298.650

^{2} = 981,110.048

which when divided by 5 = 196,222.010 which checks within the accuracy of the numbers used.

Total energy required will be 1,177,332 ev or about 1.177 Mev