# Question #f536b

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that you need to use mass of the nucleus and the combined mass of the nucleons it contains to determine the **mass defect**.

Once you know the mass defect *per nucleus*, you can use **Einstein's equation** to determine the **binding energy** *per nucleus*, then finally use Avogadro's number to find the binding energy *per mole* of nuclei.

Now, a nuclues' **mass defect** tells you the difference between the actual mass of the nucleus and the combined mass of the nucleons it contains.

A *nucleon* is a proton **or** a neutron. In your case, you know that the iron-56 nucleus has a mass of

This means that the mass defect for a single iron-56 nucleus will be

#color(blue)(M_"defect" = m_"nucleons" - m_"nucleus")#

#M_"defect" = "56.44914 u" - "55.92068 u" = "0.52846 u"#

Here's where **Einstein's equation** comes into play. The idea here is that the mass defect of a nucleus can be attributed to the **binding energy** of the nucleus, which tells you how much energy is *required* to split the nucleus into its constituent nucleons.

You can therefore, using Einstein's equation, convert the mass defect into *binding energy*

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

The trick now is to realize that you must convert the mass defect from *unified atomic mass units*, or *kilograms* - you will see why a little later on.

By definition, one unified atomic mass unit is approximately equal to

#"1 u" = 1.660539 * 10^(-27)"kg"#

This means that the mass defect *per nucleus* for iron-56 will be equal to

#0.52846color(red)(cancel(color(black)("u"))) * (1.660539 * 10^(-27)"kg")/(1color(red)(cancel(color(black)("u")))) = 8.7773 * 10^(-28)"kg"#

This means that the binding energy *per nucleus* for iron-56 will be

#E = 8.7753 * 10^(-28) * (299792458)^2 color(white)(x) overbrace("kg m"^2 "s"^(-2))^(color(red)("Joules"))#

#E = 7.8868 * 10^(-11)"J"#

Now, nuclear binding energy is usually given **per mole** of nuclei. This means that you're going to have to use **Avogadro's number** to get

#7.8868 * 10^(-11)"J"/color(red)(cancel(color(black)("nucleus"))) * (6.022 * 10^(23)color(red)(cancel(color(black)("nuclei"))))/"1 mole" = 4.749 * 10^(13)"J/mol"#

Expressed in *kilojoules per mole*, the answer will be

#4.749 * 10^(13)color(red)(cancel(color(black)("J")))/"mol" * "1 kJ"/(1000color(red)(cancel(color(black)("J")))) = color(green)(4.749 * 10^(10)"kJ/mol")#