Question:

A certain nucleus undergoes two successive decays, as shown in the decay chain below:

Given that the mass of the parent nucleus X is 100 amu and that of the daughter nucleus Y is 94 amu, determine the energy released in the first decay process, assuming the daughter nucleus Y is in its ground state.

(Assume the speed of light, c, is approximately 3.0 x 10^8 m/s)

Answer:

Let's assume that the decay process from parent nucleus X to daughter nucleus Y is an alpha decay, where an alpha particle (^4He) is emitted. In alpha decay, the nucleus X loses 2 protons and 2 neutrons. This means that the mass of the emitted alpha particle (m_alpha) is equal to 4 amu.

Given:

• Mass of parent nucleus X (m_parent) = 100 amu
• Mass of daughter nucleus Y (m_daughter) = 94 amu
• Speed of light (c) = 3.0 x 10^8 m/s

We can calculate the mass lost during the decay process using the equation:

Δm = m_parent - m_daughter - m_alpha

Substituting the given values, we have:

Δm = 100 amu - 94 amu - 4 amu

Δm = 2 amu

The energy released during the decay process can be calculated using Einstein's mass-energy equivalence equation:

E = Δm c^2

Substituting the values:

E = 2 amu * (3.0 x 10^8 m/s)^2

E ≈ 1.797 x 10^18 J

Hence, the energy released in the first decay process is approximately 1.797 x 10^18 J.