It sounds like you have a few complex physics problems related to nuclear decay and energy calculations. Let's break them down one by one to clarify the concepts and find the solutions you need.

Problem 1: Time Required to Produce 80% of Equilibrium Quantity

In this scenario, we have an unstable nuclide produced at a constant rate \( R \) through neutron absorption. The nuclide then undergoes beta decay with a half-life \( \tau \). To find the time required to produce 80% of the equilibrium quantity, we first need to understand the concept of equilibrium in this context.

Equilibrium Quantity

The equilibrium quantity \( N_{eq} \) of the nuclide can be defined as the point where the rate of production equals the rate of decay. The decay rate can be expressed as:

• Decay Rate = \( \frac{N}{\tau} \)

At equilibrium, we have:

• Production Rate = Decay Rate
• R = \( \frac{N_{eq}}{\tau} \)

From this, we can derive the equilibrium quantity:

• \( N_{eq} = R \tau \)

Finding Time for 80% of Equilibrium Quantity

To find the time required to reach 80% of \( N_{eq} \), we set up the equation:

• \( N(t) = R t - \frac{N(t)}{\tau} \)

Solving this differential equation leads us to the expression for \( N(t) \). After some calculations, we find that:

• \( N(t) = N_{eq} (1 - e^{-t/\tau}) \)

Setting \( N(t) = 0.8 N_{eq} \) and solving for \( t \) gives:

• \( t = -\tau \ln(0.2) \)

Problem 2: Kinetic Energy of the Muon

In this case, we have a meson that decays into a muon and a neutrino. Since the meson is initially at rest, we can apply conservation of momentum and energy to find the kinetic energy of the muon.

Conservation of Energy

The total energy before decay is just the rest mass energy of the meson:

• \( E_{initial} = m_{meson} c^2 \)

After the decay, the energy is distributed between the muon and the neutrino:

• \( E_{final} = m_{\mu} c^2 + K.E. + E_{neutrino} \)

By conservation of momentum, since the meson was at rest, the momentum of the muon must equal the momentum of the neutrino:

• \( p_{\mu} = p_{neutrino} \)

From these equations, we can derive the kinetic energy of the muon. The kinetic energy \( K.E. \) can be expressed as:

• \( K.E. = E_{initial} - m_{\mu} c^2 - E_{neutrino} \)

Problem 3: Speed of Daughter Nuclei and Binding Energy

For the decay of a nucleus of mass \( M + \Delta m \) into two daughter nuclei of equal mass \( M/2 \), we can again use conservation of momentum and energy.

Speed of Daughter Nuclei

Initially, the nucleus is at rest, so the total momentum is zero. After decay, if the daughter nuclei have equal mass and move in opposite directions, their momenta will cancel each other out:

• \( M_{initial} = 0 = m_{1} v_{1} + m_{2} v_{2} \)

Since \( m_{1} = m_{2} = M/2 \), we can express the speeds as:

• \( v_{1} = -v_{2} \)

Using conservation of energy, we can find the kinetic energy of the daughter nuclei and relate it to the binding energy:

• \( B.E. = \Delta m c^2 \)

For the parent nucleus, the binding energy per nucleon \( e_1 \) can be expressed as:

• \( e_1 = \frac{B.E.}{N} \)

For the daughter nuclei, the binding energy per nucleon \( e_2 \) can be similarly expressed. The relationship between \( e_1 \) and \( e_2 \) can be derived based on the mass defect before and after decay.

Final Thoughts

These problems illustrate the principles of nuclear physics, including decay processes, conservation laws, and energy transformations. If you have further questions or need clarification on any specific part, feel free to ask!