What is the nuclear binding energy of one lithium-6 atom with a measured atomic mass of 6.015 amu?

1 Answer
Dec 5, 2015


#4.890 * 10^(-12)# #"J"#


The idea here is that you need to use the mass of a single proton and the mass of a single neutron to calculate the mass of a lithium-6 nucleus, then use the measured value to find its mass defect.

As you know, a nucleus' mass defect tells you what the difference between the measured mass of the nucleus and the combined mass of the nucleons it contains.

In this case, a lithium-6 nucleus will contain

Look up the mass of a proton and the mass of a neutron

#m_"proton" = 1.6726219 * 10^(-24)"g"#

#m_"neutron" = 1.6748941 * 10^(-24)"g"#

The combined mass of the nucleons will thus be

#m_"calc" = 3 xx m_"proton" + 4 xx m_"neutron"#

#m_"calc" = 3 * 1.6726219 * 10^(-24)"g" + 3 * 1.6748941 * 10^(-24)"g"#

#m_"calc" = 1.004255 * 10^(-23)"g"#

Now, the unified atomic mass unit, #"u"#, which is defined as #1/12"th"# of the mass of a neutral carbon-12 atom in its ground state, is equivalent to

#"1 u" = 1.6605389 * 10^(-24)"g"#

This means that the measured mass of single lithium-6 atom will be equal to

#6.015 color(red)(cancel(color(black)("u"))) * (1.6605389 * 10^(-24)"g")/(1color(red)(cancel(color(black)("u")))) = 0.998814 * 10^(-23)"g"#

You can thus say that the mass defect of the lithium-6 nucleus will be equal to

#M_"defect" = m_"calc" - m_"nucleus"#

#M_"defect" = 1.004255 * 10^(-23)"g" - 0.998814 * 10^(-23)"g"#

#M_"defect" = 5.4410 * 10^(-26)"g"#

Now, mass defect and binding energy are related through Einstein's equation. The idea is that the mass defect of the nucleus can be attributed to its binding energy, which tells you how much energy is needed to split the nucleus into its constituent nucleons.

This means that you can say

#color(blue)(E = M_"defect" * c^2)#


  • #c# is the speed of light in a vacuum, equal to #2.99792458 * 10^(8)"m s"^(-1)#

Plug in your value into the above equation--do not forget to convert the mass defect from grams to kilograms, since this will allow you to express the binding energy in joules.

#E = 5.4410 * 10^(-29) * (2.99792458 * 10^8)^2color(white)(.) overbrace("kg m"^2"s"^(-2))^(color(blue)("joules"))#

#E = color(darkgreen)(4.890 * 10^(-12)"J")#

If you want, you can find the binding energy per mole of lithium-6 nuclei, which is usually expressed in kilojoules per mole.

To convert it from Joules per nucleus to kilojoules per mole, use the fact that one mole of any element contains exactly #6.022 * 10^(23)# atoms, and thus nuclei, of that element - this is known as Avogadro's number.

In your case, the binding energy for lithium-6 will be

#4.890 * 10^(-12)"J"/color(red)(cancel(color(black)("nucleus"))) * (6.022 * 10^(23)color(red)(cancel(color(black)("nuclei"))))/"1 mole" = 2.945 * 10^(12)# #"J mol"^(-1)#

Finally, this is equivalent to

#E = color(green)(2.945 * 10^9color(white)(.)"kJ mol"^(-1))#