Question

The binding energy per nucleon of $${}_3^{7}Li$$and $${}_2^{4}He$$nuclei are $5.06\mathrm{MeV}$and $7.06\mathrm{MeV}$, respectively. In the nuclear reaction$${}_3^{7}Li{+}_1^{1}H{\to }_2^{4}He{+}_2^{4}He+Q$$ What Will Be The Value Of Energy $$Q$$ released ?

Answer 1

Step 1: Given data:
The binding energy per nucleon of $${}_3^{7}Li=5.06MeV$$
The binding energy per nucleon of $${}_2^{4}He=7.06MeV$$
Also assuming the binding energy of $${}_1^{1}H$$, $$B.E_{{}_1^{1}H}=0$$

Step 2:Finding binding energy of the given nuclei :

Hence the binding energy of $${}_3^{7}Li=7\times 5.06=39.2MeV$$

And the binding energy of $${}_2^{4}He=4\times 7.06=28.24MeV$$
Step 3: Formula for Energy $$Q$$ released :

The energy released in a nuclear reaction can be given as the difference between the binding energy of reactants and products, i.e:
$$Q=B.E_{{}_2^{4}He}+B.E_{{}_2^{4}He}-B.E_{{}_3^{7}Li}-B.E_{{}_1^{1}H}$$
Where,
$$B.E_{{}_2^{4}He}$$is the binding energy Helium
$$B.E_{{}_3^{7}Li}$$is the binding energy of Lithium
$$B.E_{{}_1^{1}H}$$is the binding energy of Hydrogen
Step 4: calculation of the energy released:
Hence, the energy released during the nuclear reaction can be calculated as:

$$\begin{gathered} \mathrm{Q}=2\times \left(\mathrm{Bindingenergyof}{}_2^4\mathrm{He}\right)-\mathrm{Bindingenergyof}{}_{\mathrm{3}}^{\mathrm{7}}\mathrm{Li} \\ \Rightarrow \mathrm{Q}=2\times 28.24-7\times 5.06 \\ =17.3\mathrm{MeV} \end{gathered}$$
Hence, the Value Of Energy $$Q$$released is calculated to be $17.3\mathrm{MeV}$.