How many MeV of energy is released during a reaction in which $1.67 \times 10 ^ { - 25 } \mathrm {~kg}$ of material is converted to energy? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , 1 \mathrm { eV } = 1.67 \times 10 ^ { - 19 } \mathrm {~J} \right)$
A) $4.1 \times 10 ^ { - 7 } \mathrm { MeV }$
B) $5.0 \times 10 ^ { - 14 } \mathrm { MeV }$
C) $3.1 \times 10 ^ { - 4 } \mathrm { MeV }$
D) $9.4 \times 10 ^ { 4 } \mathrm { MeV }$

Question

Quizplus · Accepted Answer

To find the energy released, use Einstein's equation $E = mc^2$. Substituting the given values, $E = (1.67 \times 10^{-25} \, \text{kg}) \times (3.0 \times 10^8 \, \text{m/s})^2 = 1.503 \times 10^{-10} \, \text{J}$. Converting to MeV, where $1 \, \text{J} = 6.242 \times 10^{12} \, \text{MeV}$, $E = 1.503 \times 10^{-10} \, \text{J} \times 6.242 \times 10^{12} \, \text{MeV/J} = 9.4 \times 10^4 \, \text{MeV}$.

How Many MeV of Energy Is Released During a Reaction 1.67×10−25 kg1.67 \times 10 ^ { - 25 } \mathrm {~kg}1.67×10−25 kg

How Many MeV of Energy Is Released During a Reaction $1.67 \times 10 ^ { - 25 } \mathrm {~kg}$