Answer:
(a) \[E=\,m{{c}^{2}},\] where \[m=1\,u\]
\[=\,1.67\,\times \,{{10}^{-27}}\,kg\]
and \[c\,=\,3\,\times
\,{{10}^{8}}\,m{{s}^{-1}}\]
\[\therefore \] \[E\,=\,1.67\,\times
\,{{10}^{-27\,}}\times \,\,9\,\times \,{{10}^{16}}\]
\[=15.03\,\times \,{{10}^{-11}}J\]
Since \[1\,\,MeV\,=\,1.6\,\times
\,{{10}^{-13}}J\]
\[\therefore \]\[E\,=\,\frac{15.03\,\times
\,{{10}^{-11}}\,J}{1.6\,\times \,{{10}^{-13\,}}}=\,931.5\,MeV\]
(b) \[1\,u\,=\,931.5\,MeV\]
L.H.S represents mass and R.H.S
represents energy
Since \[\,m\,=\,\frac{E}{{{c}^{2}}}\]
\[\therefore
\]\[1\,u\,=\,931.5\,MeV/{{c}^{2}}\]
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