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Obtain the binding energy (in \(\mathrm{MeV}\) ) of a nitrogen nucleus \(\left({ }_{7}^{14} \mathrm{~N}\right)\), given \(m\left({ }_{7}^{14} \mathrm{~N}\right)=14,00307 \mathrm{u}\).

Short Answer

Expert verified
The binding energy of the nitrogen nucleus \({}_{7}^{14} \mathrm{~N}\) is approximately 101.11 MeV.

Step by step solution

01

Calculate Protons and Neutrons

In the nitrogen nucleus \({}_{7}^{14} \mathrm{~N}\), the atomic number (Z) is 7, indicating there are 7 protons. The mass number (A) is 14, which suggests there are \((14 - 7) = 7\) neutrons.
02

Determine Individual Masses

The mass of a proton is approximately \(1.007276\, \mathrm{u}\) and the mass of a neutron is approximately \(1.008665\, \mathrm{u}\). Therefore, the combined mass of the protons is \(7 \times 1.007276 \mathrm{u} = 7.050932 \mathrm{u}\) and for the neutrons is \(7 \times 1.008665 \mathrm{u} = 7.060655 \mathrm{u}\).
03

Calculate Total Mass of Constituents

Add the mass of protons and neutrons to get the total mass of the nucleons: \(7.050932 \mathrm{u} + 7.060655 \mathrm{u} = 14.111587 \mathrm{u}\).
04

Compute Mass Defect

The mass defect is the difference between the total mass of the individual nucleons and the actual mass of the nucleus. Therefore, mass defect \(\Delta m = 14.111587 \mathrm{u} - 14.00307 \mathrm{u} = 0.108517 \mathrm{u}\).
05

Convert Mass Defect to Energy

Use the formula \(E = \Delta m \times 931.5 \ \mathrm{MeV/u}\) to convert the mass defect into energy. \(E = 0.108517 \times 931.5 \ \mathrm{MeV/u} = 101.11 \ \mathrm{MeV}\).
06

Validate Units

Ensure all calculations use consistent units throughout. 'u' here refers to atomic mass units and \(\mathrm{MeV}\) represents mega-electronvolts. The conversion factor (931.5 \(\mathrm{MeV/u}\)) is correct for converting atomic mass units to energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nitrogen Nucleus
The nitrogen nucleus is an essential component of the nitrogen atom, symbolized as Section Tag:

  • ^{14}_{7} N,
  • with the atomic number (Z) being 7.


This signifies 7 protons in the nucleus.


Furthermore, the mass number (A) is 14, indicating that there are a total of 14 nucleons (comprising both protons and neutrons) within the nucleus. To ascertain the number of neutrons, simply subtract the atomic number from the mass number (A-Z), giving Text Tag:

  • 7 neutrons.
Together, the 7 protons and the 7 neutrons form the complete nucleus of the nitrogen atom. This nucleus is what binds the atom's electromagnetic fields, creating a unique signature that defines the nitrogen element in nature.
Mass Defect
Mass defect is a pivotal concept in nuclear physics. It represents the difference between the sum of the individual masses of protons and neutrons that historically define a nucleus, and the actual observed mass of the fully assembled nucleus.

When protons and neutrons bind together in a nucleus, some mass is converted into energy due to the binding forces according to Einstein’s equation, E=mc^{2}.

This mass difference or 'defect' plays a key role in calculating the energy that binds the nucleus together. Let's consider our example with the nitrogen nucleus. The computed mass of separate protons and neutrons amounts to approximately 14.111587 atomic mass units (u). However, when the nucleus is complete, its mass is only 14.00307 u. The mass defect ( Text Tag: r Delta m = 0.108517 u), highlights this disparity. Such mass defects are vital as they lead us to discover the binding energy, thereby revealing the stability and configuration of nuclei in atoms across the universe.
Atomic Mass Units
Understanding atomic mass units (u or amu) is essential for grasping nuclear and atomic theories. These units provide a standardized measure for expressing atomic and subatomic masses, enabling us to compare and contrast significantly different elements and isotopes. An atomic mass unit is defined primarily as one-twelfth of the mass of a carbon-12 atom. Equivalently, it is approximately equal to 1.66053906660 x 10^{-27} kg.

This precision of measurement is critical as it lets us make exact calculations when determining the properties of elements.
In nuclear equations and energy calculations, mass values in atomic mass units are often converted into energy units such as MeV (mega-electronvolts). Specifically, 1 amu is equivalent to approximately 931.5 MeV, establishing the pivotal connection between atomic mass and energy.

Utilizing atomic mass units, as shown in our exercise, allows scientists and students alike to delve into nuclear compositions and uncover deeper understandings of matter’s fundamental building blocks.

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Most popular questions from this chapter

A \(1000 \mathrm{MW}\) fission reactor consumes half of its fuel in \(5.00 \mathrm{y}\). How much \({ }_{9}^{235} \mathrm{U}\) did it contain initially? Assume that the reactor operates \(80 \%\) of the time, that all the energy generated arises from the fission of \({ }_{92}^{25} \mathrm{U}\) and that this nuclide is consumed only by the fission process.

Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius \(2.0 \mathrm{fm} .\) )

(a) Two stable isotopes of lithium \({ }_{3}^{6} \mathrm{Li}\) and \({ }_{3}^{7} \mathrm{Li}\) have respective abundances of \(7.5 \%\) and \(92.5 \% .\) These isotopes have masses \(6.01512 \mathrm{u}\) and \(7.01600 \mathrm{u}\), respectively. Find the atomic mass of lithium. (b) Boron has two stable isotopes, \({ }_{5}^{10} \mathrm{~B}\) and \({ }_{5}^{11} \mathrm{~B}\). Their respective masses are \(10.01294 \mathrm{u}\) and \(11.00931 \mathrm{u}\), and the atomic mass of boron is \(10.811 \mathrm{u}\), Find the abundances of \({ }_{5}^{10} \mathrm{~B}\) and \({ }_{5}^{11} \mathrm{~B}\).

In a periodic table the average atomic mass of magnesium is given as \(24.312 \mathrm{u} .\) The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) \((23.98504 \mathrm{u}),{ }_{12}^{25} \mathrm{Mg}(24.98584 \mathrm{u})\) and \({ }_{12}^{26} \mathrm{Mg}(25.98259 \mathrm{u})\). The naturalabundance of \({ }_{12}^{24} \mathrm{Mg}\) is \(78.99 \%\) by mass. Calculate the abundances of other two isotopes.

Suppose India had a target of producing by \(2020 \mathrm{AD}, 200,000 \mathrm{MW}\) of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was \(25 \%\). How much amount of fissionable uranium would our country need per year by \(2020 ?\) Take the heat energy per fission of \({ }^{235} \mathrm{U}\) to be about \(200 \mathrm{MeV}\).

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