Nucleus Recoil-Energy in Neutron Capture Reactions

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A nucleus which captures a thermal neutron must, since the momentum is conserved, receive a recoil energy. Immediately after capturing a neutron, the nucleus will emit γ quantas to get rid of the excess energy liberated when the neutron is bound to the nucleus. This also result in a certain amount of recoil energy on the nucleus. 

Recoil energy from n-capture

The conservation of momentum demands that

     [math]\overrightarrow{P}_n + \overrightarrow{P}_T = \overrightarrow{P}_{T+n} = \overrightarrow{P}_R[/math]

where P denotes the momentum, index n denots the neutron, index T the target nucleus, and index R the recoil. 

The general relationship between kinetic energy, EK, and momentum p is given by:

    [math]E_K = \frac{\overrightarrow{p}^2}{2m}[/math]

The mass of the neutron is 1 (atomic mass unit). the mass of the target nucleus is A. The new nucleus will therefore have mass A+1. Then

    [math]E_{K,R} = \frac{\overrightarrow{P}^2_R}{2(A+1)} = \frac{\overrightarrow{P}^2_n m_n}{2 m_n (A+1)} =\frac{E_{K,n} m_n}{A+1} =\frac{E_{K,n}}{A+1}[/math]

(remember that the momemtun of the target nucleus initially is 0.)

Recoil energy from γ emission

For emission of the mass-less quantas we have the following relationship:

    [math]\overrightarrow{P}_R = \overrightarrow{P}_\gamma[/math]


    [math]P_\gamma = \frac{E_\gamma}{c}[/math]

In this case the nucleus has mass A+1, then

    [math]E_{K,R}=\frac{\overrightarrow{P}^2_R}{2(A+1)} = \frac{\overrightarrow{P}^2_\gamma c^2}{2(A+1) c^2} = \frac{E^2_\gamma (MeV)}{2(A+1) 931.5 MeV}[/math]

The iodine case

For iodine, A = 127. Thermal neutrons have EK,n = 0.025 eV. Eγ will be around 3 MeV. We then get that

[math]E_{K,R}= \frac{0.025 eV}{128} = 0.2 meV[/math]