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Q13.1:

Obtain the binding energy (in MeV) of a nitrogen nucleus (\(^{14}_{7}N\)), given \(m (\(^{14}_{7}N\)) =14.00307 u\)

Solution :

Step 1: Identification of Nuclear Constituents

The nitrogen nucleus is denoted as $^{14}_{7}N$. Based on this notation:

  • Atomic Number ($Z$) = 7 (Number of protons)
  • Mass Number ($A$) = 14
  • Number of Neutrons ($N$) = $A - Z = 14 - 7 = 7$

Step 2: Determination of Mass Defect ($\Delta m$)

The mass defect is defined as the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus. The formula is given by: [Per the Einstein Mass-Energy Equivalence Principle]

$\Delta m = [Z \cdot m_p + N \cdot m_n] - M_{nucleus}$

Where standard values are:

  • Mass of a proton ($m_p$) = $1.007825\ u$
  • Mass of a neutron ($m_n$) = $1.008665\ u$
  • Mass of $^{14}_{7}N$ ($M_{nucleus}$) = $14.00307\ u$

Substituting the values:

$\Delta m = [7 \times 1.007825 + 7 \times 1.008665] - 14.00307$

$\Delta m = [7.054775 + 7.060655] - 14.00307$

$\Delta m = 14.11543 - 14.00307$

$\Delta m = 0.11236\ u$

Step 3: Conversion of Mass Defect to Binding Energy

To convert the mass defect in atomic mass units ($u$) to energy in MeV, we use the conversion factor: $1\ u = 931.5\ MeV/c^2$. [Per the relativistic equivalence $E = \Delta m c^2$]

$BE = \Delta m \times 931.5\ MeV/u$

$BE = 0.11236 \times 931.5$

$BE = 104.66334\ MeV$

Visual Representation of Nuclear Binding

Nucleus (7p, 7n) Binding Energy = 104.66 MeV

Calculation Summary Table

Parameter Value
Mass of 7 Protons $7.054775\ u$
Mass of 7 Neutrons $7.060655\ u$
Calculated Mass Defect $0.11236\ u$
Binding Energy $104.66\ MeV$

Final Solution: The binding energy of the $^{14}_{7}N$ nucleus is approximately 104.66 MeV.


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