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

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 $\;_{17}cl^{35}=35.3-0.3200$

$=34.98\;.$

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