MCAT Physics Question 128: Answer and Explanation

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Question: 128

8. What is the binding energy of the argon-40 isotope in MeV? (Note: m_proton = 1.0073 amu, m_neutron = 1.0087 amu, m_{Ar-40 nucleus} = 39.9132 amu,)

A. 0.4096 MeV
B. 40.3228 MeV
C. 381.7 MeV
D. 643.8 MeV

Correct Answer: C

Explanation:

To determine the binding energy, we must first determine the mass defect. The mass defect is simply the masses of each of the protons and neutrons in the unbound state added together minus the mass of the formed argon-40 nucleus (which contains 18 protons and 40 ? 18 = 22 neutrons):

The binding energy can then be determined from this mass defect:

The closest answer is choice (C).

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