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Nucleus A having \mathrm{Z=17} and equal number of protons and neutrons has \mathrm{1.2 \mathrm{MeV}} binding energy per nucleon. Another nucleus \mathrm{\mathrm{B}\: of\: Z=12} has total 26 nucleons and 1.8 \mathrm{MeV} binding energy per nucleons. The difference of binding energy of \mathrm{B \: and \: A} will be ___________\mathrm{\mathrm{MeV}.}
 

Option: 1

6


Option: 2

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Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{ \text { For Nucleus A } }
\mathrm{Z}=17=\text { Nummber of protons }           \mathrm{Given\left ( Z=N \right )\therefore N=17}
\mathrm{A}=34=\mathrm{Z}+\mathrm{N}
\mathrm{E}_{\mathrm{bn}}=1.2 \mathrm{MeV}
\frac{\mathrm{\left(\mathrm{E}_{\mathrm{B}}\right)_1 }}{A}=1.2 \mathrm{MeV}
\left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times \mathrm{A}
\left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times 34
\left(\mathrm{E}_{\mathrm{B}}\right)_1=40.8 \mathrm{MeV} \rightarrow \text { Binding energy of Nucleus } \mathrm{A}

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Deependra Verma

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