# 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"#

##### 2 Answers

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 themass defectin#"amu"# .#m_p# is the rest mass of aprotonin#"amu"# .#m_n# is the rest mass of aneutronin#"amu"# .#m_"exact"# is theexactnuclide mass in#"amu"# (including the electron(s)).#m_e# is the rest mass of anelectronin#"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")#

Li-6

MeV/atom 32.10681189

MeV/nucleon 5.348469965

#### Explanation:

Binding Energy of Li-6