What will be the recoil velocity of an electron having kinetic energy of 6 mev
Answers
Answer:
2.05.2.2.3 Energy loss and displacements produced by α-decays
A heavy recoil atom, e.g., Np237 in the decay of Am241, receives a recoil energy E due to conservation of momentum, ME = mEα, hence typically ~ 100 keV (91 keV in the decay of Am241). These recoil atoms show predominantly nuclear stopping and produce a dense collision cascade with typically ~ 1500 displacements within a short distance of ~ 20 nm. Defect clustering can occur, stabilizing the damage.
In general, ions passing through matter lose energy via two processes, either by direct collisions with the atoms of the matter (elastic collisions) or by dissipating their energy on the electrons (inelastic collisions), hence nuclear energy loss (dE/dx)n leading directly to displaced atoms, or ionizations (electronic energy loss (dE/dx)e). Fig. 7 shows the particular case of an α-decay that evidences that the heavy recoil nucleus loses its energy mainly via nuclear energy losses whereas the light α-particle does mainly by electronic energy losses. Any given radiation source can suffer these two types of energy loss. The case of α-decay in UO2 is partially included in Fig. 1 showing a displacement cascade produced by the recoil atom of the decay of U238, that is 234Th, and schematized in Fig. 7.
Fig. 7
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Fig. 7. Scheme of an α-decay event in UO2. About 100 keV of energy is dissipated by the recoil nucleus mainly elastically, whereas about 5.5 MeV are lost mainly inelastically by the α-particle, until it comes to rest and become a helium atom.
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Answer:
The energy of emitted photon, E=hf=7MeV
If p be the momentum of photon and v be the recoil velocity of nucleus, then by conservation of momentum
p=mv or E/c=mv or v=E/mc
Thus, recoil energy K=
2
mv 2=
2m ×
m c
2
E2
=
2mc
2
E 2
=
2×(24×931.5MeV)
(7MeV)
=1.09×10
−3
MeV=1.1keV
where (1amu=931.5meV)
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