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Question: Calculate the binding energy of the oxygen isotope \({ }_{8}^{16} \mathrm{O}\). The mass of the isot...

Calculate the binding energy of the oxygen isotope 816O{ }_{8}^{16} \mathrm{O}. The mass of the isotope is 16.0amu16.0 \mathrm{amu}.
(Given e=0.0005486amue=0.0005486 \mathrm{amu}, p=1.00757amup=1.00757 \mathrm{amu} and n=1.00893amun=1.00893 \mathrm{amu}).

Explanation

Solution

We know that, the binding energy is calculated by first determining the mass defect (m)\left ( \triangle m \right ), which is equal to the difference in the actual mass of nucleus and the mass of nucleus of isotopic element.

Complete step by step answer:
We can see that the symbol of oxygen isotope in the question is { }_{8}^{16} \mathrm{O}.
We know that in the symbol of an isotopic element or ion, subscript represents the atomic number and superscript represents the mass number.
Therefore, we can say that there are 88 protons, 88 neutrons and 88 electrons in 816O.{ }_{8}^{16} \mathrm{O}.
The actual mass of the nucleus of 816O=16 Mass of 8 electrons =168(0.0005486amu)=15.9956amu{ }_{8}^{16} \mathrm{O} =16-\text { Mass of } 8 \text { electrons }=16-8(0.0005486 a m u)=15.9956 a m u
The mass of the nucleus of 816O= Mass of 8 protons + Mass of 8 neutrons =8(1.00757amu)+8(1.00893amu)=16.132amu{ }_{8}^{16} \mathrm{O} = \text { Mass of } 8 \text { protons }+ \text { Mass of } 8 \text { neutrons } =8(1.00757 \mathrm{amu})+8(1.00893 \mathrm{amu})=16.132 \mathrm{amu}

Therefore, we can calculate the mass defect as follows:
Mass defect of 816O=(16.13215.9956)amu=0.1364amu{ }_{8}^{16} \mathrm{O} = (16.132-15.9956)\,{amu} = 0.1364\,{amu}
We know that the binding energy is given by (m×931MeV)\left ( \triangle m \times 931 \text {MeV} \right ).
Therefore, we can calculate the binding energy of 816O{ }_{8}^{16} \mathrm{O} as follows.
Binding energy of oxygen isotope 816O=0.1364×931MeV=127MeV.{ }_{8}^{16} \mathrm{O}=0.1364 \times 931 \mathrm{MeV}=127 \mathrm{MeV}.
Thus, we can say that the binding energy of the oxygen isotope 816O{ }_{8}^{16} \mathrm{O} is 127MeV.127 \mathrm{MeV}.

Note: We know that the superscript in the chemical symbol represents the mass number whereas the subscript denotes the atomic number of the element. It is also known that the mass number is the sum of atomic number and the number of neutrons.