Calculate (in MeV ) the total binding energy for 40 Ca.
Compute the total mass of all the individual neutrons and protons in 40Ca.
Subtract the actual mass of the nucleus of 40Ca.
Multiply the difference by c^2 (to get Joules)
Convert Joules to MeV.
To calculate the total binding energy for an atomic nucleus, you need to know the mass defect (the difference between the mass of the nucleus and the sum of the masses of its individual protons and neutrons) and use Einstein's mass-energy equivalence equation, E=mc².
1. Start by finding the atomic mass of 40Ca. The atomic mass is given in atomic mass units (u) and can be found in the periodic table. For 40Ca, the atomic mass is approximately 40.08 u.
2. Determine the number of protons and neutrons in 40Ca. Calcium (Ca) has an atomic number of 20, which means it has 20 protons. Subtracting the atomic number from the atomic mass gives the number of neutrons. Therefore, 40Ca has 40 - 20 = 20 neutrons.
3. Calculate the total mass of the protons and neutrons. The mass of a proton is approximately 1.0073 atomic mass units (u), and the mass of a neutron is approximately 1.0087 u. Multiply the number of protons by the mass of a proton and the number of neutrons by the mass of a neutron, then sum the results:
(20 protons x 1.0073 u/proton) + (20 neutrons x 1.0087 u/neutron) = approximately 40.149 u
4. Find the mass defect. Subtract the total mass of the protons and neutrons from the atomic mass of 40Ca:
40.08 u - 40.149 u ≈ -0.069 u
Since the mass defect is negative, we need to take the absolute value to make it positive (the negative sign is due to the conversion from u to energy units):
Mass defect = |-0.069 u| ≈ 0.069 u
5. Convert the mass defect to energy using Einstein's mass-energy equivalence equation:
E = mc²
where c is the speed of light, approximately 3 x 10^8 meters per second. To convert the mass defect from atomic mass units (u) to kilograms (kg), divide it by Avogadro's number (approximately 6.022 x 10²³):
0.069 u ÷ 6.022 x 10²³ u/kg ≈ 1.147 x 10⁻²⁴ kg
Now, multiply the mass by the square of the speed of light:
E = (1.147 x 10⁻²⁴ kg) x (3 x 10^8 m/s)²
E ≈ 1.034 x 10⁻⁰⁵ J
6. Convert the energy to MeV (mega-electron volts) by dividing by the electron volt-to-joule conversion factor, approximately 6.242 x 10¹⁸:
E(MeV) ≈ (1.034 x 10⁻⁰⁵ J) ÷ (6.242 x 10¹⁸ MeV/J)
E(MeV) ≈ 1.657 MeV
Therefore, the total binding energy for 40Ca is approximately 1.657 MeV.