Question

Calculate the binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\) and \({ }_{1}^{3} \mathrm{H}\). The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410\), and \({ }_{1}^{3} \mathrm{H}, 3.01605\).

Short answer

The binding energy per nucleon for deuterium (\({ }_{1}^{2} \mathrm{H}\)) is 0.8621 MeV, and for tritium (\({ }_{1}^{3} \mathrm{H}\)) is 2.6631 MeV.

Step-by-step solution

Finding the Mass Defect for Deuterium and Tritium

To find the mass defect for each isotope, we first need to determine the masses of individual protons and neutrons. The mass of a proton is approximately 1.00728 atomic mass units (amu), and the mass of a neutron is approximately 1.00867 amu. For deuterium (\({ }_{1}^{2} \mathrm{H}\)), there is 1 proton and 1 neutron in the nucleus. Therefore, the total mass of individual nucleons is: 1.00728 amu (proton) + 1.00867 amu (neutron) = 2.01595 amu The mass defect for deuterium is the difference between the mass of individual nucleons and the actual mass of the nucleus (given as 2.01410 amu): Mass Defect (Deuterium) = 2.01595 amu - 2.01410 amu = 0.00185 amu For tritium (\({ }_{1}^{3} \mathrm{H}\)), there is 1 proton and 2 neutrons in the nucleus. Therefore, the total mass of individual nucleons is: 1.00728 amu (proton) + 1.00867 amu (neutron) + 1.00867 amu (neutron) = 3.02462 amu The mass defect for tritium is the difference between the mass of individual nucleons and the actual mass of the nucleus (given as 3.01605 amu): Mass Defect (Tritium) = 3.02462 amu - 3.01605 amu = 0.00857 amu

Calculating the Binding Energy per Nucleon for Deuterium and Tritium

To calculate the binding energy per nucleon, we need to divide the mass defect for each isotope by the total number of nucleons in the nucleus and then multiply by the speed of light squared (c^2). The conversion factor between amu and MeV is 931.5 MeV/c^2. For deuterium (\({ }_{1}^{2} \mathrm{H}\)), there are 2 nucleons in the nucleus. The binding energy per nucleon is: Binding Energy per Nucleon (Deuterium) = (0.00185 amu / 2) * 931.5 MeV/c^2 = 0.000925 amu * 931.5 MeV/c^2 = 0.8621 MeV For tritium (\({ }_{1}^{3} \mathrm{H}\)), there are 3 nucleons in the nucleus. The binding energy per nucleon is: Binding Energy per Nucleon (Tritium) = (0.00857 amu / 3) * 931.5 MeV/c^2 = 0.00285667 amu * 931.5 MeV/c^2 = 2.6631 MeV

Reporting the Results

We have now calculated the binding energy per nucleon for both deuterium and tritium: Binding Energy per Nucleon (Deuterium) = 0.8621 MeV Binding Energy per Nucleon (Tritium) = 2.6631 MeV

Key concepts

Mass Defect

The mass defect is a key concept in nuclear physics and refers to the difference between the mass of an atom's protons and neutrons when they are separate compared to when they are combined in the nucleus. Due to the strong nuclear force binding these nucleons together, the combined nucleus weighs less than the sum of its parts. This is because energy is released when protons and neutrons come together, which manifests as a loss of mass according to Einstein's mass-energy equivalence principle denoted as \(E=mc^2\).

• Mass defect measures the binding energy that holds a nucleus together.
• This energy is responsible for the stability of the nucleus relating directly to nuclear binding energy.
• In the exercise, the mass defect for deuterium and tritium helps calculate their binding energy per nucleon.

By observing the mass defect, we can better understand why certain isotopes release energy during nuclear processes.

Isotopes

Isotopes are different forms of the same element, each having the same number of protons but different numbers of neutrons. This results in different atomic masses and slightly different physical properties.

• The chemical properties of isotopes are typically the same since they have the same number of electrons and protons.
• However, the nuclear properties, such as stability and binding energy, can vary significantly.
• Creating different isotopes can lead to practical applications, such as in medical imaging or nuclear energy production.

In the exercise, deuterium and tritium are isotopes of hydrogen. They have distinct mass defects and binding energies, highlighting the effects of neutron numbers on nuclear stability.

Atomic Mass Units (amu)

The atomic mass unit (amu) serves as a standard way to express atomic and molecular weights. One amu is defined as one twelfth of the mass of a carbon-12 atom, making it approximately equal to the mass of a proton or neutron.

• Atomic masses, typically presented in amu, allow comparison among different elements and isotopes.
• The exercise uses amu to calculate the mass defect of deuterium and tritium, highlighting its pivotal role in these calculations.
• Understanding atomic masses is essential as they provide a link between chemistry and physics by connecting atomic structure with nuclear binding energies.

The clarity provided by using amu units simplifies complex nuclear calculations like those involved in finding binding energies for isotopes.

Proton and Neutron Masses

Knowing the exact masses of protons and neutrons is important for calculating nuclear properties such as the mass defect and binding energy. In atomic mass units, the proton mass is approximately 1.00728 amu, while the neutron mass is roughly 1.00867 amu.

• These masses contribute to understanding the composition and stability of atomic nuclei.
• For instance, in the exercise, the precise masses allow for the calculation of mass defects in deuterium and tritium.
• Any discrepancies, like those indicated by mass defects, are transformed into binding energy, providing insight into nuclear reactions.

The knowledge of proton and neutron masses is a foundational aspect of nuclear physics, providing essential data for many nuclear calculations and energy predictions.