The mass defect for a Li-6 nucleus is -0.03434 g/mol. How do you calculate the atomic mass of Li-6?

Jun 2, 2016

6.04783 a.m.u

Explanation:

$L i - 6 \text{ }$nucleus has 3 protons and 3 neutrons.

The total mass of protons in the nucleus

$= 3 \times 1.0072766 \setminus a . m . u . \setminus = 3.02183 \setminus a . m . u$

The total mass of neutrons in the nucleus

$= 3 \times 1.0086654 \setminus a . m . u . = 3.02600 \setminus a . m . u$

The mass defect $= 0.03434 \setminus \frac{g}{m o l}$

= 0.03434 \ g/(mol.) xx (6.02xx10^23 \ " a.m.u")/g xx (1\ mol.)/(6.02xx10^23 \ "atom"

= 0.03434 \ color(red) cancel (g)/(color(green)cancel(mol.)) xx (6.02xx10^23 \ "a.m.u")/color(red) cancel(g)xx (1\ color(green)cancel(mol.))/(6.02xx10^23 \ "atom"

$= 03434 \setminus \frac{a . m . u}{\text{atom}}$

Total mass of the nucleons ( protons + neutrons)

$= 6.04783 \setminus a . m . u$

Mass of the nucleus

$= \text{Total mass of the nucleons" - "mass defect}$

$= 6.04783 \setminus a . m . u - 03434 \setminus a . m . u$

$= 6.04783 \setminus a . m . u$

Note that the mass defect is due to the binding energy - i.e. The energy required to keep the nucleons together in the nucleus.