# The binding energy of $\;_{17}cl^{35}\;$ nucleus is 298 HeV . Find its atomic mass $\;(_{1}H^{1})\;$ is 1.008143 a.m.u and that of a neutron is 1.008986 a.m.u . Given $\;1amu=931 MeV\;$

$(a)\;34.62 amu\qquad(b)\;34.98 amu\qquad(c)\;34.12 amu\qquad(d)\;34.23 amu$

Answer : (b) $\;34.98 amu$
Explanation :
The $\;_{17}cl^{35}\;$ atom has 17 protons and 18 neutrons is its nucleus
Mass of 17 protons = $\;17 \times1.008143 amu$
$=17.138 amu$
Mass of 18 neutrons= $\;18 \times 1.0089$
$=18.161 amu$
Total= 35.3 a.m.u
Mass defect $\;\bigtriangleup m=\large\frac{298}{931}=0.3201 a.m.u$
The atomic mass of $\;_{17}cl^{35}\;$ would be sum of equivalent of binding energy nucleus . Hence atomic mass of $\;_{17}cl^{35}=35.3-0.3200$
$=34.98\;.$