Respuesta :
Answer:The binding energy of the nitrogen nucleus is [tex]6.26\times 10^{28} MeV[/tex].
Explanation:
Atomic mass of the nitrogen,[tex]M_A[/tex] =14.00307 amu
Number of protons, Z = 7
Number of neutrons,N = Atomic mass - number of protons
=14.0037 - 7 =7.00307
Mass of proton ,[tex]m_p=1.00728 amu[/tex]
Mass of neutron,[tex]m_n=1.00866 amu[/tex]
[tex]\Delta m=(Zm_p+Nm_n)-M_A=(7\times 1.00728 amu+7.00307\times 1.00866 amu)-14.00307 amu[/tex]
[tex]\Delta m=0.11160 amu[/tex]
Binding energy =[tex]\Delta m\times c^2=0.11160 amu\times (3\times 10^8 m/s)^2=1.0044\times 10^{16} Joules[/tex]
[tex]1.0044\times 10^{16} Joules=6.26\times 10^{28} MeV,(1J=6.24\times 10^{12} MeV)[/tex]
The binding energy of the nitrogen nucleus is [tex]6.26\times 10^{28} MeV[/tex].
The mass defect and nuclear binding energy are equal to 0.10851 amu and 1.63*10^-11J respectively.
Data;
- mass of proton = 1.00728 amu
- mass of neutron = 1.00866 amu
- mass of nitrogen-14 atom = 1400307 amu
- mass defect ?
Mass Defect
To solve this problem, we need to find the mass defect on the atom and it can be calculated as
[tex](7*1.00728)+(7*1.00866)=14.11158amu[/tex]
Let's calculate the mass defect of this atom now
[tex]m = 14.11158-14.00307\\m = 0.10851amu[/tex]
Binding Energy
The binding energy on this atom can be calculated as
[tex]E = mc^2\\[/tex]
- m = mass defect
- c = speed of light
substitute the values and solve
[tex]E = (0.10851*1.67*10^-^2^7)*((3.0*10^8)^2\\E = 1.63*10^-^1^1J[/tex]
From the calculations above, the mass defect and nuclear binding energy are equal to 0.10851 amu and 1.63*10^-11J respectively.
Learn more on binding energy here;
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