The binding energy per nucleon of \(^{209}_{83}\mathrm{Bi}\) is ________ MeV.
\[
\text{Given: }
m\left(^{209}_{83}\mathrm{Bi}\right)=208.980388\,u,
\quad
m_p=1.007825\,u,
\quad
m_n=1.008665\,u,
\]
\[
1u = 931\,\text{MeV}/c^2
\]
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Remember:
Mass defect:
\[
\Delta m
=
(Zm_p + Nm_n) - m_{\text{nucleus}}
\]
Binding energy:
\[
\text{BE} = \Delta m \times 931\,\text{MeV}
\]
Binding energy per nucleon:
\[
\frac{\text{BE}}{A}
\]
Step 1: Understanding the Concept:
Nuclear binding energy is the energy released when a nucleus is formed from its constituent protons and neutrons.
According to Einstein's mass-energy equivalence principle (\(E=mc^2\)), this energy corresponds to a "loss" in mass, known as the mass defect (\(\Delta m\)).
The total mass of the individual nucleons is always greater than the mass of the resulting nucleus.
Binding energy per nucleon is a measure of the stability of a nucleus; higher values generally indicate more stable configurations.
To find this, we first calculate the total mass of the constituent parts, subtract the actual nuclear mass, convert the result to energy, and finally divide by the total number of nucleons (mass number \(A\)). Key Formula or Approach:
1. Mass Defect: \[ \Delta m = [Z \cdot m_p + (A-Z) \cdot m_n] - M_{nucleus} \]
2. Total Binding Energy: \[ BE = \Delta m \times 931 \text{ MeV} \]
3. BE per nucleon: \[ BE_{avg} = \frac{BE}{A} \] Step 2: Detailed Explanation:
For Bismuth-209 (\({}^{209}_{83}\text{Bi}\)):
Atomic Number (\(Z\)) = 83 (number of protons).
Mass Number (\(A\)) = 209 (total nucleons).
Number of Neutrons (\(N\)) = \(209 - 83 = 126\). 1. Calculating the theoretical mass of constituents:
Mass of protons = \(83 \times 1.007825 = 83.649475\) u.
Mass of neutrons = \(126 \times 1.008665 = 127.091790\) u.
Total constituent mass = \(83.649475 + 127.091790 = 210.741265\) u. 2. Calculating Mass Defect (\(\Delta m\)):
Actual nuclear mass = \(208.980388\) u.
\(\Delta m = 210.741265 - 208.980388 = 1.760877\) u. 3. Converting Mass to Energy:
Total Binding Energy = \(1.760877 \times 931\) MeV.
Total BE \(\approx 1639.376\) MeV. 4. Finding Binding Energy per Nucleon:
Divide the total energy by the total number of nucleons (\(A = 209\)):
\(BE/A = \frac{1639.376}{209} \approx 7.8439\) MeV.
Rounding to two decimal places gives \(7.84\) MeV. Step 3: Final Answer:
The binding energy per nucleon is \(7.84\) MeV.