L8 : Numericals - Nuclie, Physics, Class 12 Video Lecture - Class 12 | Best Video for Class 12

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8 it says obtain binding energy of the
nuclei iron 56 and Xmas 209 in units of
mega electron volts where you are given
the atomic mass of iron and the atomic
mass of bismuth so here also we have to
calculate the binding energy so for
binding energy we have to calculate the
mass defect so let us first consider
iron so what does the know how many
protons do you have for iron the atomic
number is 26 therefore number of protons
is 26 so you have 26 protons how many
neutrons mass number minus atomic number
that is 56 minus 26 which is 30 neutrons
so you have 26 protons and 30 neutrons
so what will be the mass defect mass
defect would be 26 into mass of protons
plus 30 into mass of neutron minus the
atomic mass of iron right so this will
come out to be 26 and to mass of proton
which is one point zero zero seven eight
to five plus thirty into mass of neutron
which is one point zero zero eight six
six five minus the atomic mass of iron
which is 55.93435 two eight four six one
atomic mass unit now we convert this
mass in atomic mass unit into mega
electron volts so it will become zero
point five two eight four six one into
nine thirty one point 5 mega electron
volts but C square therefore the binding
energy
we'll be equal to Delta M into C square
Delta M is zero point five two eight
four six one into nine thirty one point
5 mega electron volt per C square into C
square so the C Square and C Square will
get cancelled so this value comes out to
be four ninety two point two six mega
electron volt so this is the binding
energy of the nuclear now if I want to
calculate just for your information I'm
telling this if you want to calculate
the binding energy per nucleon what
would be that the binding energy per
nucleon will be equal to the binding
energy divided by the number of nucleon
what is number of nucleon number of
nucleon is the number of proton plus
number of neutron so that is 56 so the
binding energy per nucleon would be
equal to eight point seven six mega
electron volts so now on similar lines
you can calculate the binding energy for
bismuth because the things will remain
the same just that in case of bismuth
the number of protons would be 83 and
the number of neutrons would become 209
minus eighty three that is 126 so
otherwise everything will remain the
same so just try to calculate the
binding energy of bismuth by yourself
let's look at the fourth problem it says
Neutron separation energy is defined as
the energy required to remove a neutron
from nucleus okay so obtain the Neutron
separation energies of nuclei calcium 41
and aluminium 27 so here you are given
the atomic mass of calcium 14 and
calcium 41 and similarly you are given
the atomic mass of aluminium 26 as well
as aluminium 27 so let us first solve it
for calcium
so if you look at this calcium 41 what
happens if you take out or if you remove
a neutron from this nucleus let us
suppose we have removed one Neutron sure
that means the atomic number will remain
the same because the number of protons
are still the same
but the mass number will decrease by one
because one Neutron has been removed so
this would become your nuclear reaction
right so what is the how do we calculate
the separation energy now the separation
energy here is nothing but the binding
energy because binding energy is
actually the energy that is needed to
separate a nucleon separate the nucleons
from each other so neutron is also a
nucleon okay so now we have to calculate
the binding energy so for that we will
calculate the mass defect so mass defect
in this case will become the mass of the
constituent say the Constituent is
calcium forty and neutron so M of
calcium forty plus mass of neutron minus
the mass of calcium 41 so mass of
calcium forty is thirty nine point nine
six two five nine one plus mass of
neutron is one point zero zero eight six
six five minus mass of calcium 41 is
forty point nine six two two seven eight
so this comes out to be zero point zero
zero eight nine seven eight atomic mass
unit so now one atomic mass unit is nine
thirty one point 5 mega electron volt by
C square so using this we can convert
this mass defect also in terms of mega
electron volts it will become zero point
zero zero eight nine seven eight into
nine thirty one point 5 mega electron
volts
okay so this would become the second so
this is my Delta M so now I have to
calculate the separation energy so the
separation energy will be equal to Delta
M into C square so here it will have the
C square so this will be equal to zero
point zero zero eight nine seven eight
into nine thirty one point 5 mega
electron volts so this comes out to be
eight point three six three zero zero
seven mega electron volt now similarly
let us try to calculate the neutron
separation energy for aluminium so now
in case of aluminium also is the 1327 if
you remove one Neutron you get aluminium
1326 plus one Neutron so even in this
case the mass defect will be given by
mass of aluminium 26 plus mass of the
neutron minus mass of aluminium 27 right
so here also you put the values which
are already given in the problem so once
you put these values you get this Delta
M as 0.01 for 0x9 atomic mass unit when
you convert it into mega electron volts
you get zero point zero one four zero
one nine into 931 point 5 mega electron
volt by C square so therefore in this
case the separation energy becomes equal
to Delta M into C square which is equal
to zero point zero one four zero one
nine into 931 point 5 mega electron
volts so this value comes out to be
thirteen point zero five nine mega
electron volts
so this would be the separation Neutron
separation energy for calcium and
aluminium respectively so now you've got
an idea that how do you calculate the
binding energy of a nucleus or the
binding energy per nucleon for any atom
right okay
so now we will start with another in
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