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Câu hỏi: Cho phản ứng hạt nhân ${ }_{1}^{1} \mathrm{H}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2 \mathrm{X}$. Biết $\mathrm{m}_{\mathrm{X}}=4,0015 \mathrm{u}, \mathrm{m}_{\mathrm{H}}=1,0073 \mathrm{u}, \mathrm{m}_{\mathrm{Li}}=7,0012 \mathrm{u}$, $1 \mathrm{uc}^{2}=931 \mathrm{MeV}$ và số Avogadro $\mathrm{N}_{\mathrm{A}}=6,02 \cdot 10^{23}$. Năng lượng tỏa ra khi tổng hợp 2 g chất $\mathrm{X}$ là
A. $4.10^{23} \mathrm{MeV}$.
B. $7,7 \cdot 10^{23} \mathrm{MeV}$.
C. $15,4.10^{23} \mathrm{MeV}$.
D. $11,3 \cdot 10^{26} \mathrm{MeV}$.
A. $4.10^{23} \mathrm{MeV}$.
B. $7,7 \cdot 10^{23} \mathrm{MeV}$.
C. $15,4.10^{23} \mathrm{MeV}$.
D. $11,3 \cdot 10^{26} \mathrm{MeV}$.
$_{1}^{1}\text{H}+_{3}^{7}\text{Li}\to 2{}_{2}^{4}\text{X}$
$\Delta E=\left( {{m}_{H}}+{{m}_{Li}}-2{{m}_{X}} \right){{c}^{2}}=\left( 1,0073+7,0012-2.4,0015 \right).931=5,1205MeV$
$\dfrac{1}{2}.\dfrac{m}{{{M}_{X}}}.{{N}_{A}}.\Delta E=\dfrac{1}{2}.\dfrac{2}{4,0015}.6,{{02.10}^{23}}.5,1205\approx 7,{{7.10}^{23}}MeV$.
$\Delta E=\left( {{m}_{H}}+{{m}_{Li}}-2{{m}_{X}} \right){{c}^{2}}=\left( 1,0073+7,0012-2.4,0015 \right).931=5,1205MeV$
$\dfrac{1}{2}.\dfrac{m}{{{M}_{X}}}.{{N}_{A}}.\Delta E=\dfrac{1}{2}.\dfrac{2}{4,0015}.6,{{02.10}^{23}}.5,1205\approx 7,{{7.10}^{23}}MeV$.
Đáp án B.