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

##
#m_p = "1.00728 amu"#

#m_n = "1.00867 amu"#

#m_e = "0.000549 amu"#

I got

And just to check, this graph shows the binding energy of

If we wanted, we could get

#"32.03 MeV"/"atom" xx "1 Li-6 atom"/"6 nucleons"#

#=# #"5.338 MeV/nucleon"#

which is fairly close to the

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

CALCULATING MASS DEFECT

The mass defect is given by

#M_d = (m_p + m_n) - (m_"exact" - m_e)# ,where:

#M_d# is the mass defect in#"amu"# .#m_p# is the rest mass of a proton in#"amu"# .#m_n# is the rest mass of a neutron in#"amu"# .#m_"exact"# is the exact nuclide mass in#"amu"# (including the electron(s)).#m_e# is the rest mass of an electron in#"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_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")#

#=# #"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

I don't really remember how many

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

#M_d -= m = 0.03438 cancel"amu" xx (1/(6.0221413 xx 10^(23)) cancel"g" xx "1 kg"/(1000 cancel"g"))/cancel"amu"#

#= 5.708 xx 10^(-29)# #"kg/atom"#

CALCULATING BINDING ENERGY FROM MASS DEFECT

And so, the *binding energy* is given by:

#E = mc^2#

#= 5.708 xx 10^(-29)# #"kg"/"atom" xx (2.998 xx 10^(8) "m/s")^2#

#= 5.131 xx 10^(-12)# #"J/atom"#

And finally, in

#color(blue)(E) = (5.131 xx 10^(-12) cancel"J")/"atom" xx (cancel"1 eV")/(1.602 xx 10^(-19) cancel"J") xx "1 MeV"/(10^6 cancel"eV")#

#=# #color(blue)("32.03 MeV/atom")#

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Li-6

MeV/atom 32.10681189

MeV/nucleon 5.348469965

Binding Energy of Li-6

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The binding energy of lithium-6 is determined by calculating the mass defect and then using Einstein's mass-energy equivalence formula (E=mc^2). The mass defect is the difference between the total mass of individual nucleons and the actual mass of the nucleus.

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