Question:

A certain nuclear reaction involves the decay of thorium-230 into radium-226. The decay process can be represented as:

^230_90Th ⟶ ^226_88Ra + ^4_2He

Given that the mass of thorium-230 is 230.0496 atomic mass units (amu), the mass of radium-226 is 226.0194 amu, and the mass of a helium-4 nucleus is 4.0026 amu, calculate the energy released during this nuclear reaction in Joules (J). Assume that the speed of light, c, is approximately 3.00 x 10^8 meters per second (m/s) and use E = mc^2, where E represents energy, m represents mass, and c represents the speed of light.

Answer:

To calculate the energy released during a nuclear reaction using Einstein's mass-energy equation, we need to determine the difference in mass before and after the reaction. Let's start by calculating the mass difference:

Mass of thorium-230, m₁ = 230.0496 amu

Mass of radium-226, m₂ = 226.0194 amu

Mass of helium-4 nucleus, m₃ = 4.0026 amu

Mass difference, ∆m = (m₁ - m₂ - m₃)

∆m = (230.0496 amu - 226.0194 amu - 4.0026 amu)

∆m = 230.0496 amu - 230.0220 amu

∆m = 0.0276 amu

Next, we need to convert the mass difference to kilograms (kg) since the unit for energy is Joules (J). The conversion factor between atomic mass unit (amu) and kilograms (kg) is given by 1 amu = 1.66 x 10^-27 kg:

∆m (kg) = ∆m (amu) * (1.66 x 10^-27 kg/amu)

∆m (kg) = 0.0276 amu * (1.66 x 10^-27 kg/amu)

∆m (kg) = 4.57 x 10^-29 kg

Now, we can calculate the energy released during the nuclear reaction using Einstein's mass-energy equation, E = mc^2:

E = ∆m * c^2

E = (4.57 x 10^-29 kg) * (3.00 x 10^8 m/s)^2

E = 4.57 x 10^-29 kg * 9.00 x 10^16 m2/s2

E = 4.11 x 10^-12 J

Hence, the energy released during the nuclear reaction is approximately 4.11 x 10^-12 Joules (J).