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(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.
The proton-proton reaction that takes place in the Sun has the net effect of converting four protons (Mp= 1.007825 u) into alpha particles (Ma= 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?
binding energy:
[STRIKE]Eb=c2 * (Mnucleons-Mparticle)*931.494MeV[/STRIKE]
Eb=(Mnucleons-Mparticle)*931.494MeV/u
Eb = c2 * (4*Mp-Ma)=2.2058*1024eV =3.85408*105j
Eb*12.5% = 4.8115*104j
1*109W / Eb = 2.0783*104s-1
The correct answer is: 1.87*1021
Thank you in advance for any help anybody can provide.
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 (Mp= 1.007825 u) into alpha particles (Ma= 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]Eb=c2 * (Mnucleons-Mparticle)*931.494MeV[/STRIKE]
Eb=(Mnucleons-Mparticle)*931.494MeV/u
The Attempt at a Solution
Eb = c2 * (4*Mp-Ma)=2.2058*1024eV =3.85408*105j
Eb*12.5% = 4.8115*104j
1*109W / Eb = 2.0783*104s-1
The correct answer is: 1.87*1021
Thank you in advance for any help anybody can provide.
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