Question
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 $p+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}$ energy of proton must be(A) $28.24 \mathrm{MeV}$(B) $17.28 \mathrm{MeV}$(C) $1.46 \mathrm{MeV}$(D) $39.2 \mathrm{MeV}$
Step 1
So, the total binding energy of ${ }_{3}^{7} \mathrm{Li}$ is $7 \times 5.60 \, \mathrm{MeV} = 39.2 \, \mathrm{MeV}$. Show more…
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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 ${ }_{1}^{1} \mathrm{H}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}$ energy of proton must be (A) $39.2 \mathrm{MeV}$ (B) $28.24 \mathrm{MeV}$ (C) $17.28 \mathrm{MeV}$ (D) $1.46 \mathrm{MeV}$
If the binding energy Per nucleon in ${ }_{3}^{7} \mathrm{Li}$ and ${ }^{4}{ }_{2}$ He nuclear is $5.6 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$ respectively, then in the reaction $\mathrm{P}+{ }_{3} \mathrm{Li} \rightarrow 2\left({ }_{2}^{4} \mathrm{He}\right)$ (P here retrent Proton) energy of Proton must be (A) $1.46 \mathrm{MeV}$ (B) $39.2 \mathrm{MeV}$ (C) $17.28 \mathrm{MeV}$ (D) $28.24 \mathrm{MeV}$
The binding energies per nucleon of $\mathrm{Li}^{7}$ and $\mathrm{He}^{4}$ are $5.6 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$ respectively, then the energy of the reaction $\mathrm{Li}^{7}+p=2\left[{ }_{2} \mathrm{He}^{4}\right]$ will be (a) $17.28 \mathrm{MeV}$ (b) $39.2 \mathrm{MeV}$ (c) $28.24 \mathrm{MeV}$ (d) $1.46 \mathrm{MeV}$
Atoms, Molecules and Nuclei
Round 1
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