147453
The Binding energy per nucleon of ${ }_{3}^{7} \mathrm{Li}$ and ${ }_{2}^{4} \mathrm{He}$ nuclei are $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$, respectively.
In the nuclear reaction ${ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow 2{ }_{2}^{4} \mathrm{He}+\mathrm{Q}$, the value of energy $Q$ released is:
147454 The binding energy of deuteron $\left({ }_{1}^{2} \mathrm{H}\right)$ is $1.15 \mathrm{M}$ e $\mathrm{V}$ per nucleon and an alpha particle $\left({ }_{2}^{4} \mathrm{He}\right)$ has binding energy of $7.1 \mathrm{MeV}$ per nucleon. Then is the reaction ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{H}+\mathrm{Q}$ the energy $Q$ is
147456 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 $\mathbf{p}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}$ energy of proton must be
147453
The Binding energy per nucleon of ${ }_{3}^{7} \mathrm{Li}$ and ${ }_{2}^{4} \mathrm{He}$ nuclei are $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$, respectively.
In the nuclear reaction ${ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow 2{ }_{2}^{4} \mathrm{He}+\mathrm{Q}$, the value of energy $Q$ released is:
147454 The binding energy of deuteron $\left({ }_{1}^{2} \mathrm{H}\right)$ is $1.15 \mathrm{M}$ e $\mathrm{V}$ per nucleon and an alpha particle $\left({ }_{2}^{4} \mathrm{He}\right)$ has binding energy of $7.1 \mathrm{MeV}$ per nucleon. Then is the reaction ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{H}+\mathrm{Q}$ the energy $Q$ is
147456 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 $\mathbf{p}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}$ energy of proton must be
147453
The Binding energy per nucleon of ${ }_{3}^{7} \mathrm{Li}$ and ${ }_{2}^{4} \mathrm{He}$ nuclei are $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$, respectively.
In the nuclear reaction ${ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow 2{ }_{2}^{4} \mathrm{He}+\mathrm{Q}$, the value of energy $Q$ released is:
147454 The binding energy of deuteron $\left({ }_{1}^{2} \mathrm{H}\right)$ is $1.15 \mathrm{M}$ e $\mathrm{V}$ per nucleon and an alpha particle $\left({ }_{2}^{4} \mathrm{He}\right)$ has binding energy of $7.1 \mathrm{MeV}$ per nucleon. Then is the reaction ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{H}+\mathrm{Q}$ the energy $Q$ is
147456 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 $\mathbf{p}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}$ energy of proton must be
147453
The Binding energy per nucleon of ${ }_{3}^{7} \mathrm{Li}$ and ${ }_{2}^{4} \mathrm{He}$ nuclei are $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$, respectively.
In the nuclear reaction ${ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow 2{ }_{2}^{4} \mathrm{He}+\mathrm{Q}$, the value of energy $Q$ released is:
147454 The binding energy of deuteron $\left({ }_{1}^{2} \mathrm{H}\right)$ is $1.15 \mathrm{M}$ e $\mathrm{V}$ per nucleon and an alpha particle $\left({ }_{2}^{4} \mathrm{He}\right)$ has binding energy of $7.1 \mathrm{MeV}$ per nucleon. Then is the reaction ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{H}+\mathrm{Q}$ the energy $Q$ is
147456 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 $\mathbf{p}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}$ energy of proton must be