Solve (101) Nuclear Energy - Get pJ & MeV Answers

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The discussion focuses on calculating the energy released during the triple-alpha process, where three alpha particles combine to form carbon-12. The user seeks guidance on the necessary formulas and calculations, specifically using the mass of the alpha particles and carbon-12 to determine the energy in picojoules (pJ) and mega-electronvolts (MeV). The relevant equations include E=mc^2 and conversions between energy units. The user expresses uncertainty about the process and requests assistance in finding the solution. The conversation highlights the importance of understanding nuclear processes and energy calculations in astrophysics.
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[SOLVED] (101)Nuclear Energy

Homework Statement


According to the Big Bang model, all of the 12C in the universe was made in stars. One process that occurs in older stars is the triple-alpha process in which three alpha particles are turned into 12C. The mass of an alpha particle (an alpha particle is a 4He nucleus) is 6.646477E-27 kg. The mass of 12C is 19.926464E-27 kg. How much energy is released in the triple-alpha process? Give your answer both in pJ and in MeV.



Homework Equations


Thats.. what I need to know.
1 eV=1.6x10^-19 J
pJ= x10^-12

The Attempt at a Solution



I do not know the formula required to solve this; this topic was glanced over during a class and a half, and I had to miss one of them. Could you please help me out?

Thank you so much!
 
Physics news on Phys.org
Heard of E=mc^2 ?
Whats the mass of 3x alpha particles, what's the mass of 12c, what's the difference?
 
Haha, I feel stupid. Thank you for the quick response!
 
Can you tell me what answer you got. Thanks.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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