Nuclear Binding Energy Per Nucleon & Mass Defect Problems - Nuclear Chemistry
TLDRThis educational video script explains the concept of mass defect and nuclear binding energy per nucleon, using carbon-12 as an example. It details the steps to calculate the mass defect by comparing the mass of the nucleus to the combined mass of its constituent protons and neutrons. The script also demonstrates how to compute the nuclear binding energy using Einstein's famous equation, E = mc^2, and convert it to mega electron volts per nucleon. Additionally, it challenges viewers to apply these concepts to calculate the energy released when 5 moles of nitrogen-14 nuclei are formed, emphasizing the importance of precision in scientific calculations.
Takeaways
- π The video explains how to calculate the mass defect of an isotope and the nuclear binding energy per nucleon.
- π Mass defect is defined as the difference between the mass of the nucleus and the mass of individual particles that constitute the nucleus.
- π Carbon-12 is used as an example, with a mass number of 12 and an atomic number of 6, containing six protons and six neutrons.
- βοΈ The atomic mass unit (amu) is used to convert mass to kilograms, with 1 amu equal to 1.66054 Γ 10^-27 kilograms.
- 𧬠The mass of the carbon-12 nucleus is calculated by subtracting the mass of six electrons from the mass of the carbon atom.
- π₯ The mass of individual protons and neutrons is summed to find the total mass of the particles that make up the nucleus.
- βοΈ The mass defect of carbon-12 is calculated by subtracting the summed mass of protons and neutrons from the mass of the carbon nucleus.
- β‘ The nuclear binding energy per nucleon is found using Einstein's mass-energy equivalence formula, E=mc^2.
- π A negative energy value indicates that energy is released when the nucleus is formed, as mass is converted to energy.
- π’ The binding energy per nucleon is calculated by dividing the total binding energy by the number of nucleons in the nucleus.
- π The video also covers a problem involving the formation of nitrogen-14 nuclei from protons and neutrons, including the calculation of energy released.
Q & A
What is the mass defect of carbon-12?
-The mass defect of carbon-12 is the difference between the mass of the carbon-12 nucleus and the combined mass of its individual protons and neutrons. It is calculated to be approximately -1.643 Γ 10^-28 kilograms.
How many protons and neutrons does carbon-12 have?
-Carbon-12 has six protons and six neutrons, as the atomic number (number of protons) is six and the mass number is 12.
What is the significance of the mass of electrons in calculating the mass of the carbon nucleus?
-The mass of electrons is relatively insignificant compared to the mass of the nucleus. It can be neglected for a rough estimate, but for an exact calculation, the mass of the electrons should be subtracted from the mass of the carbon atom.
What is the conversion factor from atomic mass units (amu) to kilograms?
-One atomic mass unit (amu) is equal to 1.66054 Γ 10^-27 kilograms.
How is the nuclear binding energy per nucleon calculated?
-The nuclear binding energy per nucleon is calculated using the equation ΞE = Ξm * c^2, where Ξm is the mass defect and c is the speed of light. The result is then divided by the number of nucleons (protons and neutrons) to get the energy per nucleon.
What is the nuclear binding energy per nucleon of carbon-12 in joules and mega electron volts?
-The nuclear binding energy per nucleon of carbon-12 is approximately 1.232 Γ 10^-12 joules per nucleon. When converted to mega electron volts per nucleon, it is about 7.7 MeV/nucleon.
What is the relationship between mass defect and energy release during nuclear reactions?
-The mass defect represents the mass lost when particles come together to form a nucleus. This lost mass is converted into energy, which is released during the process. Hence, a negative energy value indicates energy release.
How many nucleons are there in a nitrogen-14 nucleus?
-A nitrogen-14 nucleus has 14 nucleons, which includes seven protons and seven neutrons.
What is the mass defect when 5 moles of nitrogen-14 nuclei are formed from protons and neutrons?
-The mass defect for the formation of 5 moles of nitrogen-14 nuclei from protons and neutrons is approximately -5.055 Γ 10^13 joules.
How is the energy released when 5 moles of nitrogen-14 nuclei are formed expressed in mega electron volts?
-The energy released when 5 moles of nitrogen-14 nuclei are formed is approximately 3.16 Γ 10^26 mega electron volts.
What is the importance of the negative sign in the calculation of energy changes in nuclear reactions?
-The negative sign in the calculation of energy changes indicates that energy is released during the process, such as when particles combine to form a nucleus. A positive value would indicate energy input, such as when a nucleus is broken apart.
Outlines
π¬ Calculating Mass Defect and Nuclear Binding Energy of Carbon-12
This paragraph introduces the concept of mass defect and nuclear binding energy, focusing on carbon-12 as an example. It explains that mass defect is the difference between the mass of the nucleus and the sum of the individual masses of protons and neutrons. The atomic mass of carbon is given as 12 amu, which converts to 1.99265 x 10^-26 kg. The mass of the carbon nucleus is calculated by subtracting the mass of six electrons from the atomic mass. The masses of protons and neutrons are also provided, and their sum is used to find the mass defect. The nuclear binding energy per nucleon is then calculated using Einstein's mass-energy equivalence formula, E=mc^2, where the mass defect is multiplied by the speed of light squared to find the energy released when the nucleus forms.
π Understanding Nuclear Binding Energy and Mass Defect Calculations
This paragraph delves deeper into the calculation of mass defect and nuclear binding energy, using the example of carbon-12. It provides the numerical values for the mass of the carbon nucleus and the individual protons and neutrons, leading to the calculation of the mass defect as -1.643 x 10^-28 kg. The nuclear binding energy is then calculated using the mass defect and the speed of light, resulting in -1.47 x 10^-11 J per nucleus. The paragraph clarifies the significance of a negative value, indicating energy release during the formation of the nucleus. It also explains how to calculate the binding energy per nucleon, which for carbon-12 is 1.232 x 10^-12 J per nucleon, and how to convert this to mega electron volts per nucleon.
π§ͺ Energy Release in Nitrogen-14 Nucleus Formation
The third paragraph discusses the process of calculating the mass defect and energy release when nitrogen-14 nuclei are formed from protons and neutrons. It begins by explaining the concept of mass defect in the context of forming a nitrogen nucleus and the expectation of energy release. The atomic mass of nitrogen is given, and the mass of the nitrogen atom is calculated by subtracting the mass of seven electrons. The masses of protons and neutrons are used to find the total mass of the subatomic particles in the nitrogen nucleus. The mass defect is calculated, and the energy released during the formation of one nitrogen nucleus is determined. The paragraph concludes with the calculation of the total energy released if five moles of nitrogen nuclei were formed.
π Total Energy Release in Nitrogen-14 Nucleus Formation Over Five Moles
This final paragraph focuses on the calculation of the total energy released when five moles of nitrogen-14 nuclei are formed. It presents the mass defect and the energy released per nucleus, then scales this up to five moles using Avogadro's number. The energy change is calculated in joules and then converted to mega electron volts, correcting a previous mistake regarding the conversion factor. The final answer is given as 3.16 x 10^26 mega electron volts, emphasizing the importance of accuracy in calculations.
Mindmap
Keywords
π‘Mass Defect
π‘Nuclear Binding Energy
π‘Carbon-12
π‘Atomic Number
π‘Neutrons
π‘Protons
π‘Electrons
π‘Atomic Mass Unit (amu)
π‘Speed of Light (c)
π‘Mega Electron Volt (MeV)
π‘Nitrogen-14
Highlights
The video explains how to calculate the mass defect of an isotope and the nuclear binding energy per nucleon.
Mass defect is the difference between the mass of the nucleus and the mass of individual particles that constitute the nucleus.
Carbon-12 has six protons and six neutrons, with the mass number being the sum of protons and neutrons.
The atomic mass unit (amu) is converted to kilograms using the conversion factor 1.66054 x 10^-27 kg.
The mass of the carbon nucleus is calculated by subtracting the mass of six electrons from the mass of the carbon atom.
The mass of the carbon atom is 1.99265 x 10^-26 kg, and the mass of its nucleus is approximately 1.99210 x 10^-26 kg.
The mass of six protons and six neutrons that make up carbon-12 is 2.00853 x 10^-26 kg.
The mass defect of carbon-12 is calculated to be -1.643 x 10^-28 kg.
Nuclear binding energy per nucleon is calculated using the equation ΞE = Ξm * c^2, where c is the speed of light.
The change in energy for carbon-12 is -1.47 x 10^-11 joules per individual nucleus, indicating energy release during nucleus formation.
The binding energy per nucleon for carbon-12 is 1.232 x 10^-12 joules per nucleon, or 7.7 mega electron volts per nucleon.
For nitrogen-14, the mass defect is calculated similarly by comparing the mass of the nucleus to the mass of individual protons and neutrons.
The mass of the nitrogen atom is 2.32527 x 10^-26 kg, and the mass of its nucleus is 2.32463 x 10^-26 kg.
The mass defect for nitrogen-14 formation from protons and neutrons is -1.8655 x 10^-28 kg.
The energy released during the formation of one nitrogen-14 nucleus is -1.679 x 10^-11 joules.
For the formation of 5 moles of nitrogen-14 nuclei, the total energy released is 5.055 x 10^13 joules or 3.16 x 10^26 mega electron volts.
The video emphasizes the importance of correctly applying the negative sign in calculations to avoid significant errors.
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