If the binding energy per nucleon in ${ }_3^7 \mathrm{Li}$ and ${ }_2^4 \mathrm{He}$ nuclei are $5.60…

If the binding energy per nucleon in ${ }_3^7 \mathrm{Li}$ and ${ }_2^4 \mathrm{He}$ nuclei are $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$ respectively, then in the reaction $\mathrm{p}+{ }_3^7 \mathrm{Li} \rightarrow 2{ }_2^4 \mathrm{He}$ energy of proton must be
  1. $39.2 \mathrm{MeV}$
  2. $28.24 \mathrm{MeV}$
  3. $17.28 \mathrm{MeV}$
  4. $1.46 \mathrm{MeV}$

Solution

$=(8 \times 7.06-7 \times 5.60) \mathrm{MeV}=17.28 \mathrm{MeV}$

Asked in: JEE Main 2006

Practice more Nuclear Physics questions on Aicharya