Uranium-238 decays into thorium-234 plus an alpha particle.The known masses are ${}^{4}_{2}$ He: 4.002603 u $^{224} _{90} Th$: 234.043583 u $^{238} _{92} U$: 238.050786 u How much energy is released in this process? (1 u = 931.5 MeV/c²)

Question

Quizplus · Accepted Answer

The energy released in the decay process can be calculated using the mass defect and converting it to energy using Einstein's equation $E=mc^2$, where $m$ is the mass defect and $c$ is the speed of light. The mass defect ($\Delta m$) is the difference between the initial mass and the final mass. Initial mass = mass of $^{238}_{92}U = 238.050786$ u Final mass = mass of $^{234}_{90}Th + {}^{4}_{2}He = 234.043583 u + 4.002603 u = 238.046186$ u Mass defect ($\Delta m$) = initial mass - final mass = $238.050786 u - 238.046186 u = 0.004600 u$ Converting the mass defect to energy: $E = \Delta m \times 931.5 \frac{MeV}{c^2}$ $E = 0.004600 u \times 931.5 \frac{MeV}{c^2} = 4.28 MeV$

Uranium-238 Decays into Thorium-234 Plus an Alpha Particle 24{}^{4}_{2}24​ He: 4002603 U

Uranium-238 Decays into Thorium-234 Plus an Alpha Particle ${}^{4}_{2}$ He: 4002603 U