Question

part a
determine the binding energy of the last neutron in a \\(_{11}^{23}\text{na}\\) nucleus. the masses of the neutral atoms of \\(_{11}^{23}\text{na}\\) and \\(_{11}^{22}\text{na}\\) are 22.989770 u and 21.984437 u, respectively. the mass of a neutron is 1.008665 u.
express your answer using four significant figures.
binding energy = \\(\boxed{}\\) mev
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Explanation:

Step1: Identify the reaction

The reaction to remove the last neutron from \(_{11}^{23}\text{Na}\) is \(_{11}^{23}\text{Na}
ightarrow _{11}^{22}\text{Na} + _0^1\text{n}\). We need to find the mass defect \(\Delta m\) for this reaction.

Step2: Calculate mass defect

The mass of \(_{11}^{23}\text{Na}\) is \(m_{23} = 22.989770\ \text{u}\), mass of \(_{11}^{22}\text{Na}\) is \(m_{22} = 21.994437\ \text{u}\), and mass of neutron \(m_n = 1.008665\ \text{u}\).

Mass defect \(\Delta m = m_{22} + m_n - m_{23}\)

Substitute the values:
\(\Delta m = 21.994437 + 1.008665 - 22.989770\)
\(\Delta m = (21.994437 + 1.008665) - 22.989770\)
\(\Delta m = 23.003102 - 22.989770\)
\(\Delta m = 0.013332\ \text{u}\)

Step3: Convert mass defect to energy

We know that \(1\ \text{u} = 931.5\ \text{MeV}\) (energy equivalent of mass).

Binding energy \(E = \Delta m \times 931.5\ \text{MeV/u}\)

Substitute \(\Delta m = 0.013332\ \text{u}\):
\(E = 0.013332 \times 931.5\)
\(E \approx 12.41\ \text{MeV}\) (rounded to four significant figures)

Answer:

\(12.41\)