The specific isotope of lithium (number of nucleons).

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The specific isotope of lithium (number of nucleons).
The mass of a neutron.
The actual atomic mass of the lithium isotope.

I will assume the most common isotope, Lithium-7 ( $^7_3\text{Li}$ ), and use standard values for the neutron mass and the atomic mass of Lithium-7.

Assumptions:
- Lithium isotope is Lithium-7 ( $^7_3\text{Li}$ ).
- Number of protons ( $Z$ ) = 3
- Number of neutrons ( $N$ ) = $7 - 3 = 4$
- Total number of nucleons ( $A$ ) = 7
- Mass of proton ( $m_p$ ) = $1.00728 \, \text{a.m.u.}$ (given)
- Mass of neutron ( $m_n$ ) = $1.00866 \, \text{a.m.u.}$ (standard value)
- Atomic mass of Lithium-7 ( $m_{\text{Li-7}}$ ) = $7.016003 \, \text{a.m.u.}$ (standard value)
- Energy equivalent of $1 \, \text{a.m.u.}$ = $931.5 \, \text{MeV}$

Calculate the total mass of the individual constituent particles (protons and neutrons).
Mass of 3 protons ( $3 \times m_p$ ):
$3 \times 1.00728 \, \text{a.m.u.} = 3.02184 \, \text{a.m.u.}$
Mass of 4 neutrons ( $4 \times m_n$ ):
$4 \times 1.00866 \, \text{a.m.u.} = 4.03464 \, \text{a.m.u.}$
Total mass of constituent particles ( $m_{\text{constituents}}$ ):
$m_{\text{constituents}} = 3.02184 \, \text{a.m.u.} + 4.03464 \, \text{a.m.u.}$ $m_{\text{constituents}} = 7.05648 \, \text{a.m.u.}$
Calculate the mass defect ( $\Delta m$ ).
The mass defect is the difference between the total mass of the individual nucleons and the actual atomic mass of the nucleus.
$\Delta m = m_{\text{constituents}} - m_{\text{Li-7}}$ $\Delta m = 7.05648 \, \text{a.m.u.} - 7.016003 \, \text{a.m.u.}$ $\Delta m = 0.040477 \, \text{a.m.u.}$
Calculate the nuclear binding energy ( $E_b$ ).
The binding energy is calculated using the mass defect and the energy equivalent of 1 a.m.u.
$E_b = \Delta m \times 931.5 \, \text{MeV/a.m.u.}$ $E_b = 0.040477 \, \text{a.m.u.} \times 931.5 \, \text{MeV/a.m.u.}$ $E_b = 37.7039555 \, \text{MeV}$
Rounding to a reasonable number of significant figures (e.g., 5, based on the input masses):
$E_b \approx 37.704 \, \text{MeV}$
Calculate the average nuclear binding energy per nucleon ( $E_{b,avg}$ ).
The average binding energy per nucleon is the total binding energy divided by the total number of nucleons ( $A$ ).
$E_{b,avg} = \frac{E_b}{A}$ $E_{b,avg} = \frac{37.7039555 \, \text{MeV}}{7}$ $E_{b,avg} = 5.386279357 \, \text{MeV/nucleon}$
Rounding to a reasonable number of significant figures:
$E_{b,avg} \approx 5.3863 \, \text{MeV/nucleon}$
The nuclear binding energy for Lithium-7 is $\boxed{\text{37.704 MeV}}$ .
The average nuclear binding energy per nucleon for Lithium-7 is $\boxed{\text{5.3863 MeV/nucleon}}$ .

The specific isotope of lithium (number of nucleons).

Answer

Step-by-step solution

Calculate the total mass of the individual constituent particles (protons and neutrons).

Calculate the mass defect ( $\Delta m$ ).

Calculate the nuclear binding energy ( $E_b$ ).

Calculate the average nuclear binding energy per nucleon ( $E_{b,avg}$ ).

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The specific isotope of lithium (number of nucleons).

Answer

Step-by-step solution

Calculate the total mass of the individual constituent particles (protons and neutrons).

Calculate the mass defect (Δm\Delta mΔm).

Calculate the nuclear binding energy (EbE_bEb​).

Calculate the average nuclear binding energy per nucleon (Eb,avgE_{b,avg}Eb,avg​).

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Still stuck on this one?

More Chemistry Questions

Calculate the mass defect ( $\Delta m$ ).

Calculate the nuclear binding energy ( $E_b$ ).

Calculate the average nuclear binding energy per nucleon ( $E_{b,avg}$ ).