How Is Binding Energy Calculated for Lithium-6 Nuclei?

[Physics Help!]

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.

 

[technician]

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?

 

[Physics Help!]

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!

 

[technician]

that is good to hear ! sometimes we are blinded by information!

 

[asteriatic]

I can provide some guidance on how to find the binding energy for the nuclei of Lithium 6. The equation you are using is correct, but there may be some mistakes in your calculations. Here are a few things to consider:

1. Make sure you are using the correct values for the atomic mass and the masses of the particles involved. The atomic mass of Lithium 6 is actually 6.0151223 u, which may make a small difference in your final answer.

2. The units you are using are also important. In your equation, you are using the mass of a nucleon in kilograms, but the binding energy per nucleon is given in MeV. Make sure all your units are consistent.

3. When using the equation for binding energy, it is important to consider the mass defect. This is the difference between the actual mass of the nucleus and the sum of the masses of its individual particles. In your equation, you are using the mass of the Lithium 6 nucleus (6.015 u), but this already includes the mass defect. You do not need to subtract it again.

Taking these factors into consideration, the correct calculation should give you a binding energy of approximately 31.9 MeV. This is slightly different from the given value of 5.33 MeV per nucleon, but this could be due to rounding errors or slight variations in the values used.

I hope this helps clarify the calculation for you. If you have any further questions, feel free to ask. Keep up the good work in your scientific endeavors!