Question

Packing fraction, an informative method previously used to decide stability of a nucleus is given by :
$Packingfraction=\frac{{}_Z^A\text{M}-A}{A}\times {10}^4$, where ${}_Z^A\text{M}$ is actual mass of nuclide in amu and $A$ is mass number. Which statements are correct about packing fraction?

• A. A nuclide with negative packing fraction has an atomic mass ${}_Z^A\text{M}$ less than its mass number
  $e.g.{,}_{28}^{56}\text{M}{(}_Z^A\text{M}55.9349amu)$
• B. A nuclide with zero packing fraction has an atomic mass ${}_Z^A\text{M}$ equal to its mass number
  $e.g.{,}_8^{16}\text{O}{(}_Z^A\text{M}16amu)$
• C. A nuclide with positive packing fraction has an atomic mass ${}_Z^A\text{M}$ more than its mass number
  $e.g.{,}_1^2\text{H}{(}_Z^A\text{M}=2.014amu)$
• D. More is the magnitude of negative packing fraction, more will be stability of nucleus

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A A nuclide with negative packing fraction has an atomic mass ${}_Z^A\text{M}$ less than its mass number
$e.g.{,}_{28}^{56}\text{M}{(}_Z^A\text{M}55.9349amu)$
B A nuclide with positive packing fraction has an atomic mass ${}_Z^A\text{M}$ more than its mass number
$e.g.{,}_1^2\text{H}{(}_Z^A\text{M}=2.014amu)$
C A nuclide with zero packing fraction has an atomic mass ${}_Z^A\text{M}$ equal to its mass number
$e.g.{,}_8^{16}\text{O}{(}_Z^A\text{M}16amu)$
D More is the magnitude of negative packing fraction, more will be stability of nucleus

$Packingfraction\left(P\right)=\frac{{}_Z^A\text{M}-A}{A}\times {10}^4$, where ${}_Z^A\text{M}$
so
$P<0,whenM<A$
$P=0,whenM=A$
$P>0,whenM>A$
More is the magnitude of negative packing fraction, more will be stability of nucleus as it has already loosen all its extra energy.