AP Physics 2 Exam Question:

Consider a nuclear reaction between two isotopes of helium: helium-3 (He-3) and helium-4 (He-4). The reaction can be represented as follows:

He-3 + He-4 → H-1 + H-1 + H-1

Where H-1 represents a hydrogen nucleus (a proton).

1. Given that the mass of He-3 is 3.01605 atomic mass units (u), the mass of He-4 is 4.00260 u, and the mass of H-1 is 1.00783 u, calculate the energy released in this nuclear reaction.

Answer:

To calculate the energy released in this nuclear reaction, we can use the concept of mass-energy equivalence, given by Einstein's famous equation:

E = mc²

Where E is the energy, m is the mass, and c is the speed of light (approximately 3.00 x 10^8 m/s).

Step 1: Calculate the total mass of the reactants (He-3 and He-4): Total mass of reactants = mass of He-3 + mass of He-4 = 3.01605 u + 4.00260 u = 7.01865 u

Step 2: Calculate the total mass of the products (H-1, H-1, and H-1): Total mass of products = 3 × mass of H-1 = 3 × 1.00783 u = 3.02349 u

Step 3: Calculate the mass defect (∆m) in the reaction: ∆m = Total mass of reactants - Total mass of products = 7.01865 u - 3.02349 u = 3.99516 u

Step 4: Calculate the energy released (∆E) using mass-energy equivalence: ∆E = ∆m × c² = 3.99516 u × (3.00 x 10^8 m/s)² ≈ 3.59564 x 10^-11 kg × (3.00 x 10^8 m/s)² ≈ 3.23607 x 10^-2 J

The energy released in this nuclear reaction is approximately 3.23607 x 10^-2 J.

Note: The mass difference (∆m) in the reaction corresponds to the mass converted into energy, which is the energy released during the nuclear reaction.