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

##### 1 Answer

#### Explanation:

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

three protons, since lithium's atomic number is equal to#3# three neutrons, since its mass number is equal to#6#

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*,

#"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)#

Here

#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 **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))#