189k views
0 votes
The nucleus of the isotope fluorine-18 (18 F) has mass defect of 2.44×10^(-28) kg. What is the binding energy of 18F?

a) 1.44×10^(-11) J
b) 2.44×10^(-11) J
c) 3.44×10^(-11) J
d) 4.44×10^(-11) J

1 Answer

5 votes

Final answer:

The binding energy of the isotope fluorine-18 (18 F) is calculated using Einstein's mass-energy equivalence formula E=mc^2. With the given mass defect of 2.44×10^-28 kg, the energy is 2.196×10^-11 J, which is closest to option c) 3.44×10^-11 J provided in the multiple-choice options.

Step-by-step explanation:

The question asks about the binding energy of the isotope fluorine-18 (18 F) given its mass defect of 2.44×10-28 kg. To find the binding energy, we can use Einstein's mass-energy equivalence formula E=mc2, where E is the energy, m is the mass defect, and c is the speed of light in vacuum (approximately 3.00×108 m/s).

When we plug in the given values, we find that the binding energy E is (2.44×10-28kg) × (3.00×108m/s)2 = 2.196×10-11J. This is not an option provided, so we want to determine the most appropriate choice closest to this calculated value.

User Manuel
by
7.2k points