Four Protons Are Fused Into An Alpha Particle, Energy Released

[breakingaway]

(answered)Four Protons Are Fused Into Alpha Particle, Energy Released

Answered: I wasn't supposed to multiply by c². Thank you ehild

I apologize in advance if I posted this in an incorrect format or in the wrong section. I read the rules and I believe this is the correct format/place.

Homework Statement

The proton-proton reaction that takes place in the Sun has the net effect of converting four protons (M_p= 1.007825 u) into alpha particles (M_a= 4.002603 u). The resulting "released" energy is shared between the kinetic energy of the alpha particles, positrons, neutrinos and gamma ray photons. Assume we could eventually design a 1.00 GW (billion watt) fusion power plant using this same reaction. Also assume that 12.5% of the "released" energy (kinetic + photon) can be converted into electrical power.
How many alpha particles would need to be fused in the reactor every second?

Homework Equations

binding energy:
[STRIKE]E_b=c² * (M_nucleons-M_particle)*931.494MeV[/STRIKE]
E_b=(M_nucleons-M_particle)*931.494MeV/u

The Attempt at a Solution

E_b = c² * (4*M_p-M_a)=2.2058*10²⁴eV =3.85408*10⁵j

E_b*12.5% = 4.8115*10⁴j

1*10⁹W / E_b = 2.0783*10⁴s^-1

The correct answer is: 1.87*10²¹

Thank you in advance for any help anybody can provide.

 

[ehild]

In what units of mass have you calculated with? ehild

 

[breakingaway]

In what units of mass have you calculated with? ehild

For the problem the mass of a proton(1.007825u) and alpha particle(4.002603u) is given in atomic mass units(u=931.494MeV). I converted to joules before doing any calculations involving power.

 

[ehild]

Why did you multiplied energy by c^2?

ehild

 

[breakingaway]

Why did you multiplied energy by c^2?

ehild

Well, I think you just found my mistake.

Just crunched the number without the c^2 and it came out correct.

Thank you.

 

[ehild]

Yes, taking the proton mass in kg-s, m(proton)c²/(1.6 10^-19(J/eV) )≈931 MeV.

ehild