How do I find the binding energy for the nuclei of Lithium 6.
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.
To find the binding energy of the nucleus of Lithium 6, you are using the correct equation. However, there seems to be an error in the values you are using for the masses of the particles.
Let's go through the calculation step by step:
1. Find the mass of a neutron (mass of n):
The mass of a neutron is approximately 1.008664 atomic mass units (u).
2. Find the mass of Lithium 6 (m(N,Z)):
The mass of Lithium 6 is approximately 6.015 atomic mass units (u).
3. Calculate the binding energy using the formula you provided:
B(N,Z) = Z * m(0,1) * c^2 + N * (mass of n) * c^2 - m(N,Z) * c^2
Here, Z = 3 (number of protons in the nucleus) and N = 3 (number of neutrons in the nucleus). Also, c is the speed of light, which is approximately 3 × 10^8 m/s.
Now, let's plug in the correct values and calculate the binding energy:
B(N,Z) = 3 * (1.00783 u * 1.66E-27 kg) * (3E8 m/s)^2 + 3 * (1.008664 u * 1.66E-27 kg) * (3E8 m/s)^2 - (6.015 u * 1.66E-27 kg) * (3E8 m/s)^2
Calculating this expression will give you the correct binding energy of the Lithium 6 nucleus.