Physics · Composition and size of nucleus, atomic masses, mass-energy relation, mass defect
Let be the mass of proton, the mass of a neutron, the mass of a nucleus, and the
Let \( m_{p} \) be the mass of proton, \( m_{n} \) the mass of a neutron, \( M_{1} \) the mass of a \( _{10}^{20} N e \) nucleus, and \( M_{2} \) the mass of a \( _{20}^{40} C a \) nucleus. Then This question has multiple correct options
- A. \( M_{2}=2 M_{1} \)
- B. \( M_{2}>2 M_{1} \)
- C. \( M_{2}<2 M_{1} \)
- D. \( M_{1}<10\left(m_{p}+m_{p}\right) \)
Step-by-step solution
The mass of a nucleus is less than the sum of its constituent nucleons due to binding energy. For Ne-20 (10 protons, 10 neutrons) and Ca-40 (20 protons, 20 neutrons), the binding energy per nucleon is higher for Ca-40. Therefore, the total binding energy of Ca-40 is more than twice that of Ne-20, leading to a larger mass defect. Consequently, the actual mass M2 of Ca-40 is less than twice the mass M1 of Ne-20, i.e., M2 < 2M1.
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