Answer:
Here, \[{{E}_{1}}=14.4keV=14.4\times
1.6\times {{10}^{-16}}J\], \[{{M}_{2}}=56.935\], \[u=56.935\times
1.66\times {{10}^{-27}}kg\]
According to the principle of conservation of linear momentum
\[\frac{{{E}_{1}}}{c}+\sqrt{2{{M}_{2}}{{E}_{2}}}=0\] \[\therefore
\] \[-\frac{{{E}_{1}}}{c}=\sqrt{2{{M}_{2}}{{E}_{2}}}\] or \[\frac{{{E}_{1}}^{2}}{{{c}^{2}}}=2{{M}_{2}}{{E}_{2}}\]
\[\therefore \] \[{{E}_{2}}=\frac{E_{1}^{2}}{2{{M}_{2}}{{c}^{2}}}=\frac{{{\left(
14.4\times 1.6\times {{10}^{-16}} \right)}^{2}}}{2\times 56.935\times
1.66\times {{10}^{-27}}{{\left( 3\times {{10}^{8}} \right)}^{2}}}J\]
\[{{E}_{2}}=3.1185\times
{{10}^{-22}}J=\frac{3.1185\times {{10}^{-22}}}{1.6\times
{{10}^{-16}}}keV=1.95\times {{10}^{-6}}keV\]
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