AP Physics 2 Exam Question: Nuclear Reactions
A nuclear reaction occurs when the nucleus of an atom undergoes a change in its composition, resulting in the emission of radiation or the formation of a new atom. Consider a nuclear reaction in which a radium-226 nucleus decays into an element X by the emission of an alpha particle. The alpha particle consists of two protons and two neutrons.
Using the information provided, write down the equation representing the nuclear reaction of radium-226 decaying into element X by the emission of an alpha particle.
Calculate the mass defect, Δm, for the nuclear reaction given that the mass of a radium-226 nucleus is 226.0254 atomic mass units (amu). The mass of the emitted alpha particle is 4.0015 amu, and the mass of the resulting element X is 222.0176 amu. Show all your work.
Calculate the change in energy, ΔE, for the nuclear reaction given that the speed of light, c, is approximately 3.00 × 10^8 meters per second. Express your answer in joules (J). Show all your work.
Answer with Step-by-Step Explanation:
^226/88Ra -> ^222/86X + ^4/2He
Δm = (mass of radium-226 nucleus) - (mass of emitted alpha particle + mass of resulting element X)
Δm = 226.0254 amu - (4.0015 amu + 222.0176 amu) Δm = 226.0254 amu - 226.0191 amu Δm = 0.0063 amu
Therefore, the mass defect (Δm) for the nuclear reaction is 0.0063 amu.
ΔE = Δmc²
Where Δm is the mass defect calculated in part 2, and c is the speed of light.
Converting the mass defect to kilograms: Δm_kg = (0.0063 amu) x (1.66 × 10^(-27) kg / 1 amu) Δm_kg = 1.0458 × 10^(-29) kg
Substituting the values into the equation for ΔE: ΔE = (1.0458 × 10^(-29) kg) x (3.00 × 10^8 m/s)² ΔE = 9.4136 × 10^(-13) J
Therefore, the change in energy (ΔE) for the nuclear reaction is 9.4136 × 10^(-13) Joules.