Get Answers to all your Questions

header-bg qa

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


Option: 2


Option: 3


Option: 4


Answers (1)


\mathrm{ \text { For Nucleus A } }
\mathrm{Z}=17=\text { Nummber of protons }           \mathrm{Given\left ( Z=N \right )\therefore N=17}
\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}

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE