Nuclear Physics - Mass Defect & Binding Energy

AI Thread Summary
The discussion focuses on calculating the mass defect for Carbon 12 using the provided equation. The nuclear mass of Carbon 12 is given as 1.99264 x 10^-26 kg, while the masses of protons and neutrons are also specified. A participant inquires about the mass of an electron, which is crucial for an accurate calculation. It is emphasized that the total mass of 12 electrons, approximately 0.01093 x 10^-27 kg, should not be overlooked due to its significance in the final result. Accurate consideration of all particle masses is essential for determining the mass defect correctly.
Kylah
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[SOLVED] Nuclear Physics - Mass Defect & Binding Energy

1. Carbon 12 (^{12}_{6}C) has a nuclear mass of 1.99264 x 10-26 kg, a proton has a mass of 1.67353 x 10-27, and a neutron has a mass of 1.67492 x 10-27 kg. Calculate the mass defect for carbon 12.

My equation looks like this:
\Deltam = [Z(mp+me) + (A-Z)mn]-matom

Where:
\Deltam = ?
mp = 1.67353 x 10-27 kg
mn = 1.67353 x 10-27 kg
me =
matom = 1.99264 x 10-26
Z = 6
A = 12


What I'm not sure is what me is. Could anybody help me?
 
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Hey,

The mass of an electron is 1836 times smaller than the mass of a proton m_{e}=9.10938*10^{-31}Kg, though I think that even if you take it into consideration in your calculation you will find it makes no difference to your result, since your other numbers don't have sufficient decimal places.
 
Actually the electron mass will be significant here. 12 electrons will have a total mass of 0.01093 x 10^-27kg. With 6 significant digits, this mass can not be ignored.
 
Chi Meson said:
Actually the electron mass will be significant here. 12 electrons will have a total mass of 0.01093 x 10^-27kg. With 6 significant digits, this mass can not be ignored.

Thank you!
 
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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