Question

Find the binding energy of the tritium nucleus: [Given: mass of $ ^3H = 3.01605 \, u $; $ m_p = 1.00782 \, u $; $ m_n = 1.00866 \, u $]

• A. 8.5 MeV
• B. 8.5 J
• C. 0.00909 MeV
• D. 0.00909 eV

Hint:

The binding energy can be calculated using the mass defect, which involves subtracting the mass of the nucleus from the sum of the masses of its constituent nucleons. Always remember to convert mass defect to energy using \(E = mc^2\).

Answer 1

The Correct Option is A

The binding energy of a nucleus is given by the mass defect formula: \[ E_b = (\Delta m) \times c^2 \] Where: - \(E_b\) is the binding energy, - \(\Delta m\) is the mass defect, - \(c\) is the speed of light (\(c = 3 \times 10^8 \, \text{m/s}\)). The mass defect \(\Delta m\) is the difference between the total mass of the separated nucleons and the mass of the nucleus: \[ \Delta m = [Z m_p + (A - Z) m_n - m_{\text{nucleus}}] \] For tritium (\(^3H\)): - \(Z = 1\) (number of protons), - \(A = 3\) (mass number), - \(m_{\text{nucleus}} = 3.01605 \, u\), - \(m_p = 1.00782 \, u\), - \(m_n = 1.00866 \, u\). Substitute these values into the mass defect formula: \[ \Delta m = [1 \times 1.00782 + (3 - 1) \times 1.00866 - 3.01605] \] \[ \Delta m = [1.00782 + 2 \times 1.00866 - 3.01605] \] \[ \Delta m = [1.00782 + 2.01732 - 3.01605] \] \[ \Delta m = 0.00909 \, u \] Now, convert the mass defect to kilograms by multiplying with the atomic mass unit in kilograms (\(1 \, u = 1.66 \times 10^{-27} \, \text{kg}\)): \[ \Delta m = 0.00909 \, u \times 1.66 \times 10^{-27} \, \text{kg} = 1.51 \times 10^{-29} \, \text{kg} \] The energy in joules is: \[ E_b = \Delta m \times c^2 = (1.51 \times 10^{-29}) \times (3 \times 10^8)^2 \] \[ E_b = 1.36 \times 10^{-12} \, \text{J} \] Now, to convert to MeV, use the conversion factor \(1 \, \text{J} = 6.242 \times 10^{12} \, \text{MeV}\): \[ E_b = 1.36 \times 10^{-12} \times 6.242 \times 10^{12} = 8.5 \, \text{MeV} \] Thus, the binding energy is \(8.5 \, \text{MeV}\).