Question

How do you calculate the mass defect for C-14? How do you calculate the binding energy?

Answer 1

Well, you take the theoretical mass of combining all the particles, and subtract from it the actual isotopic mass. That mass accounts for the binding energy.

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You'll need:

• `m_p = "1.00727647 amu/proton"`
• `m_n = "1.00866492 amu/neutron"`
• `m_e = 5.48530 xx 10^(-4) "amu/electron"`
• `m_(""^(14) "C") = "14.003241989 amu/atom"`
• There are `("1.60217662 J")/("1 eV")`.
• The speed of light is `2.99792458 xx 10^8 "m/s"`.
• Avogadro's number is `6.0221413 xx 10^23 "mol"^(-1)`.

We know that `""_6^(14) "C"` has `6` protons and `8` neutrons (and `6` electrons), so the theoretical mass (that is, if no mass became released energy due to forming the atom) is:

`"6 protons" cdot "1.00727647 amu/proton" + "6 electrons" cdot 5.48530 xx 10^(-4) "amu/electron" + "8 neutrons" cdot "1.00866492 amu/neutron"`

`=` `"14.11626936 amu/atom"`

The mass defect is then

`color(blue)(m_"defect") = m_"theoretical" - m_(""^(14) "C")`

`= "14.11626936 amu/atom" - "14.003241989 amu/atom"`

`=` `color(blue)("0.11302737 amu")`

`"amu"` is numerically the same as `"g/mol"`, so we can then get to the binding energy. This binding energy is found as follows...

The mass that becomes energy is:

`m = (0.11302737 cancel"g")/cancel"mol" xx "1 kg"/(1000 cancel"g") xx (cancel"1 mol")/(6.0221413 xx 10^(23))`

`= 1.87686348 xx 10^(-28) "kg"`

So, the binding energy involved in `"J"` per atom is:

`E_"binding" = mc^2`

`= (1.87686348 xx 10^(-28) "kg")(2.99792458xx10^(8) "m/s")^2`

`= 1.68680000^(-11) "J/atom"`

We may want it in `"kJ/mol"` and `"MeV/nucleon"`.

`color(blue)(E_"binding") = (1.68680000 xx 10^(-11) cancel"J")/cancel"atom" xx "1 kJ"/(1000 cancel"J") xx (6.0221413 xx 10^23 cancel"atoms")/("1 mol")`

`= color(blue)(1.015815 xx 10^(10) "kJ/mol")`

`color(blue)(E_"binding") = (1.68680000 xx 10^(-11) cancel"J")/cancel"atom" xx cancel"1 eV"/(1.60217662 xx 10^(-19) cancel"J") xx ("1 MeV")/(10^6 cancel"eV") xx cancel"1 C-14 atom"/("14 nucleons")`

`=` `color(blue)("7.520127 MeV/nucleon")`

The actual value is `"7.520319 MeV/nucleon"`.