A nucleus of mass 218 amu is in free state decays to emit an alpha particle.Kinetic energy of alpha particle emitted is 6.7Mev.The recoil energy of daughter nucleus in Mev is

To find the recoil energy of the daughter nucleus after the emission of an alpha particle, we can use the principles of conservation of momentum and energy. When the alpha particle is emitted, the total momentum of the system must remain constant. Let's break this down step by step.

Understanding the Decay Process

In this scenario, we have a nucleus with a mass of 218 atomic mass units (amu) that decays by emitting an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, which gives it a mass of approximately 4 amu. After the emission, the remaining nucleus, known as the daughter nucleus, will have a mass of:

• Mass of daughter nucleus = Mass of parent nucleus - Mass of alpha particle
• Mass of daughter nucleus = 218 amu - 4 amu = 214 amu

Applying Conservation of Momentum

Before the decay, the total momentum of the system is zero since the nucleus is at rest. After the decay, the momentum of the alpha particle and the daughter nucleus must also sum to zero. If we denote the velocity of the alpha particle as \( v_{\alpha} \) and that of the daughter nucleus as \( v_d \), we can express this as:

• Momentum of alpha particle + Momentum of daughter nucleus = 0
• m_α * v_α + m_d * v_d = 0

Where:

• m_α = mass of alpha particle = 4 amu
• m_d = mass of daughter nucleus = 214 amu

Calculating the Velocities

From the momentum conservation equation, we can express the velocity of the daughter nucleus in terms of the alpha particle's velocity:

• v_d = - (m_α / m_d) * v_α

Relating Kinetic Energy to Velocity

The kinetic energy (KE) of the alpha particle is given as 6.7 MeV. The kinetic energy can also be expressed in terms of mass and velocity:

• KE_α = (1/2) * m_α * v_α²

We can rearrange this to find the velocity of the alpha particle:

• v_α = sqrt((2 * KE_α) / m_α)

Finding the Recoil Energy of the Daughter Nucleus

The kinetic energy of the daughter nucleus can be calculated using its velocity:

• KE_d = (1/2) * m_d * v_d²

Substituting for \( v_d \) from our earlier equation:

• KE_d = (1/2) * m_d * \[(- (m_α / m_d) * v_α)\]²

Now, substituting \( v_α \) and simplifying, we can find the recoil energy of the daughter nucleus:

• KE_d = (1/2) * m_d * \[(m_α / m_d)^2 * (2 * KE_α / m_α)\]

Final Calculation

Plugging in the values:

• m_α = 4 amu
• m_d = 214 amu
• KE_α = 6.7 MeV

After performing the calculations, we find:

• KE_d = (1/2) * 214 * \[(4 / 214)^2 * (2 * 6.7 / 4)\]

After simplifying, the recoil energy of the daughter nucleus comes out to be approximately 0.062 MeV.

Thus, the recoil energy of the daughter nucleus is around 0.062 MeV, which is significantly less than the kinetic energy of the emitted alpha particle due to the large mass difference between the alpha particle and the daughter nucleus.