Question

The binding energy per nucleon for magnesium-27 is $\mathbf{1}\mathbf{.}\mathbf{326}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{nucleon}$. Determine the atomic mass of magnesium-27.

Short answer

The atomic mass of magnesium-27 is 26.9830 amu.

Step-by-step solution

Atomic mass.

In simple words, atomic mass refers to the mass of an atom. Atomic mass is expressed as an atomic mass unit (AMU). The mass of both elements and molecules is found by atomic mass.

Using the Einstein mass energy relation.

The binding energy per nucleon for magnesium$-27=1.326\times {10}^{-12}\mathrm{J}$ /nucleon.

The following method will be used to calculate the energy in joules:

$\begin{array}{c}\mathrm{\Delta E}=1.326\times {10}^{-12}\frac{\mathrm{J}}{\mathrm{nucleon}}\times 27\mathrm{nucleon}\\ =3.5802\times {10}^{-11}\mathrm{J}\end{array}$

The Einstein mass energy will be used to calculate the mass defect-

$\Delta E=\Delta m.c^2..........\left(1\right)$

Where,

• $\Delta E=$Change in energy
• $\Delta m=$Change in mass
• $c^2=$Velocity of light

Atomic mass of magnesium-27 .

Put the given values in (1)

$\begin{array}{c}3.5802\times {10}^{-11}J=\Delta m\times {\left(3.00\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2\\ \Delta m=3.978\times {10}^{28}kg\end{array}$

Now, convert the kg into amu.

$\begin{array}{c}\Delta m=3.978\times {10}^{-28}\mathrm{kg}\times \frac{6.072\times {10}^{23}\mathrm{amu}}{1\mathrm{kg}}\\ \Delta m=0.23955\mathrm{amu}\end{array}$

Magnesium-27 has 12protons and 5 neutrons, so the equation becomes:

$\begin{array}{c}0.2399amu=m_{Mg}-\left(12m_p+15m_n\right)\\ 0.2399amu=m_{Mg}-\left[12\left(1.00728amu\right)+15\left(1.00866amu\right)\right]\\ m_{Mg}=26.9764amu\end{array}$

The total mass including the mass of 12 electrons is calculated.

$\begin{array}{c}Massof{}^6\text{Li}atom=26.9764amu+12m_e\\ =26.9764amu+12\left(5.49\times {10}^{-4}amu\right)\\ =26.9830amu\end{array}$