The energy released when $\dfrac{7}{17.13}$ kg of $^7_3\text{Li}$ is converted into $^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32}$ eV. The value of $\alpha$ is ______. (Nearest integer) (Mass of $^7_3\text{Li} = 7.0183$ u, mass of $^4_2\text{He} = 4.004$ u, mass of proton $= 1.008$ u and $1\text{u} = 931\ \text{MeV/c}^2$ and Avogadro number $= 6.0 \times 10^{23}$)
Numerical value type. Enter your answer.
Answer: 6
Concept. Find $Q$ for one reaction $^7\text{Li}+p\to2\,{}^4\text{He}$, then multiply by the number of lithium nuclei.
Step 1. $Q$. $\Delta m=7.0183+1.008-2\times4.004=0.0183$ u, so $Q=0.0183\times931=17.0$ MeV.
Step 2. Number of nuclei. $\dfrac{7000/17.13\ \text{g}}{7\ \text{g mol}^{-1}}\times6\times10^{23}=3.50\times10^{25}$.
Step 3. Energy. $3.50\times10^{25}\times17.0\times10^6=5.97\times10^{32}$ eV.
Answer. $\alpha=6$.
Solution by Sreeraj P, M.Sc Physics
