Problem 27 - A Complete Resource Book in Chemistry for JEE Main

Question

What will be the binding energy of \(\mathrm{O}^{16}\), if the mass defect is \(0.210\) amu? (a) \(1.89 \times 10^{10} \mathrm{~J} \mathrm{~mol}^{-1}\) (b) \(1.89 \times 10^{12} \mathrm{~J} \mathrm{~mol}^{-1}\) (c) \(1.89 \times 10^{13} \mathrm{~J} \mathrm{~mol}^{-1}\) (d) \(1.89 \times 10^{14} \mathrm{~J} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The binding energy of \(\mathrm{O}^{16}\) is \(1.89 \times 10^{13} \text{ J/mol}\), answer (c).

Step 1: Understanding Mass Defect

The mass defect is the difference between the mass of a nucleus and the sum of the individual masses of its protons and neutrons. It is given as \(0.210\) amu for \(\mathrm{O}^{16}\).

Step 2: Calculate Energy Using Einstein’s Equation

To find the binding energy, use Einstein’s equation for mass-energy equivalence: \[ E = \Delta m \times c^2 \]where \(\Delta m\) is the mass defect and \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\).First, convert the mass defect from amu to kg:\[ 1 \text{ amu} = 1.660539 \times 10^{-27} \text{ kg} \]\(0.210 \text{ amu} = 0.210 \times 1.660539 \times 10^{-27} \text{ kg}\) \(\approx 3.487132 \times 10^{-28} \text{ kg}\) Now calculate \(E\):\[ E = \Delta m \times c^2 = 3.487132 \times 10^{-28} \text{ kg} \times (3.00 \times 10^8 \text{ m/s})^2 \]\[ E \approx 3.1391188 \times 10^{-11} \text{ J} \text{ per nucleus} \]

Step 3: Convert to Energy per Mole

1 mole of \(\mathrm{O}^{16}\) contains \(6.022 \times 10^{23}\) nuclei (Avogadro's number). The total binding energy for 1 mole is:\[ E_{\text{mole}} = 3.1391188 \times 10^{-11} \text{ J/nucleus} \times 6.022 \times 10^{23} \text{ nuclei/mol} \]\[ \approx 1.8899 \times 10^{13} \text{ J/mol} \]

Key Concepts

Mass Defect

The concept of mass defect is crucial in understanding the binding energy of a nucleus. Mass defect refers to the discrepancy between the mass of an intact nucleus and the sum of the individual masses of the neutrons and protons that constitute it. This discrepancy arises because a certain amount of mass is converted into energy, which holds the nucleus together. In the example of \(\mathrm{O}^{16}\), the mass defect is provided as \(0.210\) amu (atomic mass units). To make calculations manageable, we often need to convert this mass defect from amu to kilograms since energy calculations using Einstein's equation require mass in standard units of kilograms. Calculating the mass defect helps us determine how much energy is being "lost" or converted during the formation of a nucleus, which is crucial for calculating the binding energy.

Einstein’s Equation

One of the most famous equations in physics, Einstein’s Equation, \( E = \Delta m \times c^2 \), connects mass and energy directly. In this equation:

• \( E \) stands for energy
• \( \Delta m \) is the mass defect in kilograms
• \( c \) is the speed of light, which is a constant \( 3.00 \times 10^8 \text{ m/s} \)

Using this equation, the mass defect converts to energy, demonstrating the profound equivalence between mass and energy. For the case of \(\mathrm{O}^{16}\), we first convert the mass defect into kilograms and then apply the equation to get the energy in joules. This energy value represents how much energy per nucleus is required to hold the nucleus together, giving insight into the stability and strength of the bonds within the nucleus.

Avogadro's Number

Avogadro's Number serves a fundamental role in chemistry and physics. It is defined as \(6.022 \times 10^{23}\), indicating how many units (like atoms or molecules) are present in one mole of a substance. This number allows us to translate microscopic measurements into macroscopic quantities we can measure and observe. When calculating binding energy for a mole of a substance, Avogadro's Number helps us scale up from the energy of a single nucleus to the energy for a full mole of nuclei. With this, we ensure calculations reflect energy scales that are meaningful in practical, laboratory, or industrial settings. By multiplying the energy per nucleus by Avogadro's Number, we obtain the total binding energy per mole, providing a more accessible figure for understanding and comparisons.