When a stationary plutonium-239, $^{239}_{94}$ Pu, decays into uranium-235 plus an alpha particle, the energy released in the process is 5.24 MeV. The following masses are known: $^{4}_{2}$He: 4.002603 u $^{235}_{92}$U: 235.043924 u What is the mass of the $^{239}_{94}$ Pu nucleus, in amu? (1 u = 931.494 MeV/c²)

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The correct answer is A.The mass of the $^{239}_{94}$ Pu nucleus can be calculated using the following equation:$$Mass\ of\ ^{239}_{94}Pu = Mass\ of\ ^{235}_{92}U + Mass\ of\ ^{4}_{2}He + Energy\ released$$$$Mass\ of\ ^{239}_{94}Pu = 235.043924\ u + 4.002603\ u + 5.24\ MeV/c^2$$$$Mass\ of\ ^{239}_{94}Pu = 235.043924\ u + 4.002603\ u + 5.24\ MeV/c^2 * (1\ u/931.494\ MeV/c^2)$$$$Mass\ of\ ^{239}_{94}Pu = 235.043924\ u + 4.002603\ u + 0.005622\ u$$$$Mass\ of\ ^{239}_{94}Pu = 239.05215\ u$$

When a Stationary Plutonium-239 94239^{239}_{94}94239​ Pu, Decays into Uranium-235 Plus an Alpha Particle, the Energy

When a Stationary Plutonium-239 $^{239}_{94}$ Pu, Decays into Uranium-235 Plus an Alpha Particle, the Energy