Energy from alpha decay, physics problem...?
Calculate the energy (in MeV) released when α decay converts polonium 210Po (atomic mass = 209.982857 u) into lead 206Pb (atomic mass = 205.974449 u). The atomic mass of an α particle is 4.002603 u.
Calculate the energy (in MeV) released when α decay converts polonium 210Po (atomic mass = 209.982857 u) into lead 206Pb (atomic mass = 205.974449 u). The atomic mass of an α particle is 4.002603 u.
Comments
1st determine the total mass of the products.
205.974449 u + 4.002603 u = 209.977052 u
The mass of the products is less than the mass of the reactant, which is weird.
Where did the mass go?
Let’s determine the difference of the final and initial mass.
209.982857 u – 209.977052 u = 0.0005805 u
The letter, u, represents the atomic mass unit, amu.
1 atomic mass unit = 931.5 Mev
Total energy released = 0.0005805 * 931.5 ≈ 5.407 Mev
The mass did not become energy. The neutrons and protons just became more stable. As objects become more stable, they have less inertia. Mass actually measures the inertia of the atoms of an object.
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convinced, darealbud is nice, it does use conservation of momentum - yet i imagine we may be able to simplify it down slightly: seeing that momentum is conserved, the momentum of the alpha particle is an same because the momentum of the daughter nucleus, merely interior the different route. a million. enable 0 characterize the alpha particle, and a million characterize the daughter nucleus: m0 * v0 = -m1 * v1 2. Kinetic power E = (a million/2)*(m)*(v^2), so: v = sqrt(2*E/m) 3. Substituting again into the momentum equation: m0 * sqrt(2*E0/m0) = -m1 * sqrt(2*E1/m1) 4. sq. both area: m0^2 * (2*E0/m0) = m1^2 * (2*E1/m1) 5. once you cancel out variables, you could finally get: (m0 / m1) * E0 = E1 because you comprehend m0, m1 and E0, you'll get E1. desire this facilitates! ^_^