# How do you determine the binding energy of lithium-6?

## ${m}_{p} = \text{1.00728 amu}$ ${m}_{n} = \text{1.00867 amu}$ ${m}_{e} = \text{0.000549 amu}$

Jul 10, 2017

I got $\text{32.03 MeV/atom}$.

And just to check, this graph shows the binding energy of $\text{_(3)^(6) "Li}$ in $\text{MeV/nucleon}$.

If we wanted, we could get

$\text{32.03 MeV"/"atom" xx "1 Li-6 atom"/"6 nucleons}$

$=$ $\text{5.338 MeV/nucleon}$

which is fairly close to the $5.3$ or so $\text{MeV/nucleon}$ on the graph.

The binding energy is the energy that is needed to hold the nucleus together, and accounts for the small mass defect found in certain nuclides.

If we find the mass defect, i.e. how much less mass the nuclide has compared to the expected mass of the protons + neutrons, then it turns out that mass incorporated into $E = m {c}^{2}$ gives the binding energy.

CALCULATING MASS DEFECT

The mass defect is given by

${M}_{d} = \left({m}_{p} + {m}_{n}\right) - \left({m}_{\text{exact}} - {m}_{e}\right)$,

where:

• ${M}_{d}$ is the mass defect in $\text{amu}$.
• ${m}_{p}$ is the rest mass of a proton in $\text{amu}$.
• ${m}_{n}$ is the rest mass of a neutron in $\text{amu}$.
• ${m}_{\text{exact}}$ is the exact nuclide mass in $\text{amu}$ (including the electron(s)).
• ${m}_{e}$ is the rest mass of an electron in $\text{amu}$.

Now, by recalling that the number of neutrons + protons is given by the mass number, and the number of protons by the atomic number, we have that ${m}_{n} = A - Z$, and:

M_d = [overbrace(3)^(Z) xx overbrace("1.00728 amu")^"Mass of one proton" + (overbrace(6)^(A) - overbrace(3)^(Z)) xx overbrace("1.00867 amu")^"Mass of one neutron"] - ("6.01512 amu" - 3 xx overbrace("0.000549 amu")^"Mass of one electron")

$=$ $\text{0.03438 amu}$

Now, what we'll need to do to get this to work with the Einstein relationship is to convert the mass to $\text{kg}$, since a $\text{J}$ is a ${\text{kg"cdot"m"^2"/s}}^{2}$, and there are about $1.602 \times {10}^{- 19}$ $\text{J/eV}$.

I don't really remember how many $\text{g/amu}$ there are, but I do remember that if we represent Avogadro's number as ${N}_{A}$, then with ${N}_{A} = 6.0221413 \times {10}^{23}$ ${\text{mol}}^{- 1}$, we have that$\frac{1}{N} _ A = \text{number of g/amu}$.

So, we have the per-atom mass defect as:

${M}_{d} \equiv m = 0.03438 \cancel{\text{amu" xx (1/(6.0221413 xx 10^(23)) cancel"g" xx "1 kg"/(1000 cancel"g"))/cancel"amu}}$

$= 5.708 \times {10}^{- 29}$ $\text{kg/atom}$

CALCULATING BINDING ENERGY FROM MASS DEFECT

And so, the binding energy is given by:

$E = m {c}^{2}$

$= 5.708 \times {10}^{- 29}$ "kg"/"atom" xx (2.998 xx 10^(8) "m/s")^2

$= 5.131 \times {10}^{- 12}$ $\text{J/atom}$

And finally, in $\text{MeV/atom}$, just like "megabytes", there are one million $\text{eV}$ in one $\text{MeV}$. So, we would then have:

$\textcolor{b l u e}{E} = \left(5.131 \times {10}^{- 12} \cancel{\text{J")/"atom" xx (cancel"1 eV")/(1.602 xx 10^(-19) cancel"J") xx "1 MeV"/(10^6 cancel"eV}}\right)$

$=$ $\textcolor{b l u e}{\text{32.03 MeV/atom}}$

Jul 10, 2017

Li-6
MeV/atom 32.10681189
MeV/nucleon 5.348469965

#### Explanation:

Binding Energy of Li-6