How Is Binding Energy Calculated for Lithium-6 Nuclei?

AI Thread Summary
The discussion focuses on calculating the binding energy for Lithium-6 nuclei, with participants addressing the formula used and the correct interpretation of given values. The initial calculation yielded a binding energy of 30.2 MeV, which was incorrect, leading to confusion about the binding energy per nucleon. After clarification, it was confirmed that using the binding energy per nucleon (5.33 MeV) multiplied by the number of nucleons (6) should yield the total binding energy. The corrected calculation resulted in a binding energy of approximately 31.98 MeV, aligning with the expected outcome. The conversation highlights the importance of understanding the relationship between binding energy and the number of nucleons in nuclear physics.
Physics Help!
Messages
12
Reaction score
0

Homework Statement


How do I find the binding energy for the nuclei of Lithium 6.
N=Z=3, atomic mass=6.015 u, binding energy per nucleon= 5.33 MeV
The equation I'm using is :

B(N,Z)=Z*m(0,1)*c^2+N*(mass of n)*c^2-m(N,Z)*c^2.

B(N,Z)=3*(1.00783u*1.66E-27kg)*(3E8 m/s)^2+3*(1.008664u*1.66E-27kg)*(3E8 m/s)^2-(6.015u**1.66E-27kg)*(3E8 m/s)^2

B(N,Z)= 4.84056E-12 J *6.24E18 eV= 30.2 MeV

but that's not the answer, I also tried dividing that answer by the binding energy per nucleon which is 5.332 MeV and I got 5.66 MeV that's not the answer. I'll appreciate it if someone explained to me what I'm doing wrong. Thanks.
 
Physics news on Phys.org
You are given the binding energy per nucleon and you know the number of nucleons so you should be able to get the binding energy of the NUCLEUS.
Is that what the question means?
 
Well, you are right, I didn't catch that , now I got 31.98 MeV, which is close to the same answer I got doing all the needless calculations but not quite, since now I got it right. Thanks!
 
that is good to hear ! sometimes we are blinded by information!
 
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}...
Back
Top