Question

A neutron breaks into a proton and an electron. This decay of neutron is accompanied by the release of energy. Assuming that 50% of the energy is produced in the form of electromagnetic radiation, what will be the frequency of radiation produced. Will this photon be sufficient to cause ionization of aluminum? In case it is able to do so what will be the energy of the electron ejected from the aluminum atom.

$I{E}_1$ of $Al=$ 577 KJ/mol

Answer 1

$n\to P^{+}+e^{-}E=0.782MeV$

a) Half of energy is in form of electromagnetic radiation=$\frac{0.782}{2}=0.391MeV$

$E_{emw}=hv\Rightarrow 0.391\times {10}^6\times 1.6\times {10}^{-19}=6.626\times {10}^{-34}\times vv=9.4\times {10}^{19}Hz$

b)Ionization energy of Al per atom=$=\frac{577}{6.023\times {10}^{23}}=9.58\times {10}^{-22}kJ=9.58\times {10}^{-19}J=5.9875eV$

Since $E_{ionisation}<E_{emw}$

Therefore the energy released in neutron decay can easily ionised Aluminium.