Question

How much energy must be supplied to break a single aluminum-27 nucleus into separated protons and neutrons if an aluminum-27 atom has a mass of 26.9815386 amu? (the mass of an electron is 5.485799×10−4 amu, the mass of a proton is 1.0072765 amu, and the mass of a neutron is 1.0086649 amu.) express your answer using six significant figures. G?

Answer 1

Solution: For calculating nuclear binding energy we need to find the binding energy and that can be calculated by mass defect. Mass defect is the excess mass present in any isotope.

$_{13}^{27}\textrm{Al}$ nucleus having 13 protons as it is visible from atomic number and number of neutrons is 14 which is calculated from $\text{atomic mass= number of neutrons + number of protons}$.

As it is given in the question that $m_p= 1.0072765amu$ and $m_n=1.0086649amu$, we can calculate the mass of the isotope.

$\text{Mass of Isotope}= (13\times1.0072765)amu+(14\times1.0086649)amu$

$\text{Mass of isotope}=27.2159031amu$

Mass defect or $\Delta m$ = mass of isotope - atomic mass.

$\Delta m=27.2159031-26.9815386$

$\Delta m=0.2343645amu$

As the mass is expressed in kilograms do to convert it into kilograms, we need to use the conversion $(1amu=1.6606\times10^-^2^7kg)$

$\Delta m=(0.2343645\frac{amu}{nucleus})(1.6606\times10^-^2^7\frac{kg}{amu})$

$\Delta m=3.891856\times10^-^2^6kg/nucleus$

To calculate energy binding energy we will use the formula

$\Delta E=\Delta mc^2$

where c= speed of light $(2.9979\times10^8m/s)$

$\Delta E=(3.891856\times10^-^2^6kg/nucleus)(2.9979\times10^8m/s)^2$

$\Delta E=3.497768\times10^-^1^1 J/nucleus$

Nuclear binding energy is expressed in kJ/mol of nuclei, so for that we need to convert Joules to kJ by conversion $(1kJ=1000J)$ and converting individual particles into moles, we need to multiply it by avagadro's number that is $6.022\times10^2^3nuclei/mol$.

$\text{Nuclear Binding Energy}= (3.497768\times10^-^1^1\frac{J}{nucleus})(\frac{1kJ}{1000J})(6.022\times10^2^3\frac{nuclei}{mol})$

$N.B.E.=2.106355\times10^1^1\text{kJ/mol of nuclei}$

Amount of energy to break down the energy of a single bond of  $_{13}^{27}\textrm{Al}$ nuclei is $2.106355\times10^1^1kJ/\text{mol of nuclei}$