Calculating Nuclear Binding Energy For Lithium Atom

Understanding nuclear binding energy is crucial in the realm of nuclear physics. This energy represents the amount of energy required to disassemble a nucleus into its constituent protons and neutrons, or conversely, the energy released when these nucleons combine to form a nucleus. The concept is rooted in the famous mass-energy equivalence principle articulated by Albert Einstein, $E=mc^2$, where a small amount of mass (the mass defect) corresponds to a substantial amount of energy. In this article, we will delve into calculating the nuclear binding energy for a lithium atom, utilizing the provided mass defect and the speed of light.

The nucleus of an atom is composed of protons and neutrons, collectively known as nucleons. The mass of a nucleus is slightly less than the sum of the masses of its individual nucleons. This difference in mass, known as the mass defect, is a manifestation of the energy that binds the nucleons together. When nucleons come together to form a nucleus, energy is released, and the system transitions to a lower energy state, resulting in a decrease in mass. This energy, the nuclear binding energy, is what holds the nucleus together against the electrostatic repulsion between the positively charged protons. Calculating the nuclear binding energy involves determining the mass defect and then using Einstein's equation, $E=mc^2$, to convert this mass difference into energy.

The mass defect, denoted as $\Delta m$, can be calculated using the following formula:

$\Delta m = (Z \times m_p + N \times m_n) - m_{nucleus}$

Where:

• $Z$ is the number of protons.
• $m_p$ is the mass of a proton.
• $N$ is the number of neutrons.
• $m_n$ is the mass of a neutron.
• $m_{nucleus}$ is the actual mass of the nucleus.

Once the mass defect is known, the nuclear binding energy ($E$) can be calculated using Einstein's equation:

$E = \Delta m \times c^2$

Where:

• $c$ is the speed of light in a vacuum ($approx 3 \times 10^8 m/s$).

This equation shows that even a small mass defect can result in a significant amount of energy due to the large value of $c^2$. The nuclear binding energy is typically expressed in units of joules (J) or megaelectronvolts (MeV). Understanding and calculating nuclear binding energy is essential for analyzing nuclear reactions, nuclear stability, and the energy released in nuclear processes.

The problem at hand requires us to calculate the nuclear binding energy for a lithium atom. We are given the mass defect, which is $6.9986235 \times 10^{-29}$ kilograms. We are also provided with the value for the speed of light, $c = 3 \times 10^8 m/s$. The formula to be used is Einstein's mass-energy equivalence equation, $E = mc^2$.

Step-by-Step Calculation

1. Identify the Given Values

We are given:

• Mass defect ($m$) = $6.9986235 \times 10^{-29}$ kg
• Speed of light ($c$) = $3 \times 10^8$ m/s

2. Apply Einstein's Equation

The equation to calculate the nuclear binding energy ($E$) is:

$E = mc^2$

3. Substitute the Values

Substitute the given values into the equation:

$E = (6.9986235 \times 10^{-29} kg) \times (3 \times 10^8 m/s)^2$

4. Perform the Calculation

First, square the speed of light:

$(3 \times 10^8 m/s)^2 = 9 \times 10^{16} m^2/s^2$

Next, multiply the mass defect by the square of the speed of light:

$E = (6.9986235 \times 10^{-29} kg) \times (9 \times 10^{16} m^2/s^2)$

$E = 6. 29876115 \times 10^{-12} kg \cdot m^2/s^2$

Since 1 Joule (J) is equivalent to 1 $kg \cdot m^2/s^2$:

$E = 6.29876115 \times 10^{-12} J$

5. Round to Significant Figures

The given answer options are provided with two significant figures, so we round our result accordingly:

$E \approx 6.3 \times 10^{-12} J$

Detailed Discussion of the Result

The calculated nuclear binding energy for the lithium atom is approximately $6.3 \times 10^{-12}$ joules. This value represents the energy released when the nucleons (protons and neutrons) come together to form the lithium nucleus, or the energy required to break the nucleus apart into its constituent nucleons. The magnitude of this energy highlights the tremendous forces at play within the nucleus, which are far stronger than typical chemical bond energies.

To put this value into perspective, consider that typical chemical reactions involve energy changes on the order of electron volts (eV) or kilojoules per mole (kJ/mol). In contrast, nuclear reactions involve energies on the order of megaelectronvolts (MeV), which are millions of times greater. The nuclear binding energy of $6.3 \times 10^{-12}$ joules, when converted to MeV, demonstrates the immense energy associated with nuclear processes.

To convert joules to MeV, we use the conversion factor:

$1 MeV = 1.602 \times 10^{-13} J$

So,

$E (in MeV) = \frac{6.3 \times 10^{-12} J}{1.602 \times 10^{-13} J/MeV} \approx 39.3 MeV$

This result indicates that the nuclear binding energy for the lithium atom is approximately 39.3 MeV, a substantial amount of energy that underscores the stability of the nucleus and the strong nuclear force that holds it together. This energy is a direct consequence of the mass defect, which in this case, is the small amount of mass that is converted into energy according to Einstein's equation.

The concept of nuclear binding energy is pivotal in understanding nuclear stability and the energy released in nuclear reactions. Nuclei with higher binding energies per nucleon are more stable, and the energy released in nuclear fission and fusion processes is directly related to the differences in binding energies between the initial and final nuclei. The accurate calculation and interpretation of nuclear binding energies are, therefore, essential in nuclear physics and related fields.

In conclusion, we have successfully calculated the nuclear binding energy for a lithium atom using the given mass defect and the speed of light. By applying Einstein's mass-energy equivalence equation, $E = mc^2$, we determined the binding energy to be approximately $6.3 \times 10^{-12}$ joules. This result underscores the immense energy associated with nuclear forces and the stability of atomic nuclei. Understanding nuclear binding energy is crucial for further studies in nuclear physics and related applications.