In a radioactive decay, $2.37 \times 10^{-3} \mathrm{u}$ are converted into kinetic energy. How much kinetic energy is this? ( $1 \mathrm{u}$ $=1.66 \times 10^{-27} \mathrm{~kg})$
A) $3.93 \times10^{30} \mathrm{~J}$
B) $2.13 \times 10^{14} \mathrm{~J}$
C) $3.54 \times 10^{-13} \mathrm{~J}$
D) $9.31 \times 10^{8} \mathrm{~J}$
E) $1.60 \times 10^{-13} \mathrm{~J}$

Question

In a radioactive decay, $2.37 \times 10^{-3} \mathrm{u}$ are converted into kinetic energy. How much kinetic energy is this? ( $1 \mathrm{u}$ $=1.66 \times 10^{-27}  \mathrm{~kg})$
A) $3.93 \times10^{30} \mathrm{~J}$
B) $2.13 \times 10^{14}  \mathrm{~J}$
C) $3.54 \times 10^{-13}  \mathrm{~J}$
D) $9.31 \times 10^{8} \mathrm{~J}$
E) $1.60 \times 10^{-13} \mathrm{~J}$

Quizplus · Accepted Answer

The mass-energy equivalence principle, given by Einstein's equation $E=mc^2$, where $m$ is the mass in kilograms, $c$ is the speed of light in vacuum ($3.00 \times 10^8 \, \text{m/s}$), can be used to calculate the kinetic energy produced from the mass converted during radioactive decay. First, convert the mass from unified atomic mass units (u) to kilograms: $2.37 \times 10^{-3} \, \text{u} \times 1.66 \times 10^{-27} \, \text{kg/u} = 3.9342 \times 10^{-30} \, \text{kg}$. Then, apply the equation $E=mc^2$: $E = 3.9342 \times 10^{-30} \, \text{kg} \times (3.00 \times 10^8 \, \text{m/s})^2 = 3.54 \times 10^{-13} \, \text{J}$.

In a Radioactive Decay 2.37×10−3u2.37 \times 10^{-3} \mathrm{u}2.37×10−3u Are Converted into Kinetic Energy

In a Radioactive Decay $2.37 \times 10^{-3} \mathrm{u}$ Are Converted into Kinetic Energy