L7 : Numericals - Nuclie, Physics, Class 12 Video Lecture - Class 12 | Best Video for Class 12

hello friends this video on nuclei part
7 is brought to you by exam fee or form
no move here from exam please make sure
that you have watched all the videos Tim
pad 6 before going ahead with pad 7 so
let us now quickly look at some of the
problems based on whatever we studied so
far the first problem see is that the
three stable isotopes of neon that is
neon 20 neon 21 and new on 22 respective
abundances of 90.5 1% 0.25 percent and
9.2 2% the atomic masses of the three
isotopes are 19.99 atomic mass unit
twenty point nine nine atomic mass unit
and twenty one point nine nine atomic
mass unit respectively obtain the
average atomic mass of neon okay so what
is given here three isotopes of neon are
given and these three isotopes have got
same atomic number because they are all
neon but their mass numbers are
different since their mass numbers are
different therefore their masses are
also different so let us suppose let us
denote the atomic mass of each of these
isotopes as M 1 M 2 and M 3 similarly
let us denote their abundances as N 1 n
2 and n 3 now the problems is that if
you have it you think it in this way let
us suppose you have a sample you have a
sample of or you have a solution okay
that's why you should is made up of that
is nothing but a mixture of water oil
and let us say water oil and say
carousel so these are the three liquids
which are mixed together to form a
liquid now I say that water is present
in the abundance of water is 20%
the abundancy of
as the 60% and the abundancy of oil is
again 20% right so similar is the
situation here so when I say neon neon
has different isotopes right they can
exist in different forms so it says that
these isotopes of neon their abundances
are in this much percentages so they are
present in this much percentages and the
atomic masses of these three isotopes
are given so you have to calculate the
average atomic mass of neon so what
would be the average atomic mass that
would be nothing but an one percentage
of m1 is present that means n-1
percentage of this much mass is there
again this much percentage of m2 is
there again this much percentage of m3
is there right this is how they are
present there if I say that 2 percent of
neon 20 is present there and mass of
neon 20 is 2 grams that means you have 2
percent of 2 grams present there so that
is what I have done here in what
percentage of m1 right now you have to
calculate the average weight so you
divide it by n 1 plus n 2 plus n 3 right
so this now we can put the values what
is N 1 N 1 is given as 90.5 1 into M 1 M
1 is given as 19.99
plus M 2 M 2 that is twenty point nine
nine into zero point two seven plus
twenty one point nine nine into nine
point two two this divided by n 1 plus n
2 plus n 3 that is equal to 90 point
five one plus zero point two seven plus
nine point two two
so this comes out to be twenty point one
seven seven one atomic mass unit so this
would be the average atomic mass of neon
right so this is the formula which which
we normally use to calculate average of
anything okay
I mean suppose if I say you have got you
obtain 30 in your physics 40 in your
chemistry and 20 in biology so what is
the average mass you obtained in science
so then also you will do the same thing
right okay so this is how we do it the
total sum the sum divided by the total
number of samples so here you had this
abundances in how many percentage each
of them is present so it is present in
ninety point five one percent so that
means ninety point five one percent of
its mass right because this neon is
present in this much percentage so that
means this much percentage of its mass
will contribute to the average mass so
therefore ninety point five one percent
off 19.99 19.99 is its mass similarly
what would be the second one's
contribution it would be zero point two
seven percent off its mass so zero point
five seven percent of its mass similarly
the third one and since we want to
calculate the average atomic mass so we
divided it by the amount of their event
I mean the amount in which they are
present in the sample so that is n1 plus
n2 plus n3 clear okay
so let us go ahead and look at the next
problem it says obtain the binding
energy in mega electron volts of a
nitrogen nucleus n 714 given the mass of
nitrogen as fourteen point zero zero
three zero seven atomic mass unit so how
do we calculate the binding energy so
when I talk about nitrogen so what is
the composition of nitrogen it has seven
protons and seven neutrons
like so that means mass of seven protons
plus mass of seven neutrons actually
forms this nitrogen right now in order
to calculate the binding energy we will
have to calculate the mass defect
because binding energy is given as Delta
M into C square where Delta n is the
mass defect so let us first calculate
mass defect so how do we define mass
defect mass defect is the sum of the
masses of the constituents so these are
the constituents minus the nuclear mass
of the atom so what is the mass of the
constituents here it will be seven into
mass of protons plus seven into mass of
neutrons this is the mass of the
constituents minus the atomic mass of
nitrogen right so here we are neglecting
the mass of electron because mass of
electron is very less so this we can
write it as seven into mass of proton
what is the mass of proton it is one
point zero zero seven eight two five
atomic mass unit plus 7 into one point
zero zero eight six six five atomic mass
units minus the mass of nitrogen that is
14 point zero zero three zero seven so
this comes out to be zero point one one
to these things atomic mass unit so this
is my mass defect now we have to
calculate in terms of mega electron
volts so that means we have to establish
a relationship between atomic mass unit
and mega electron volt you remember when
I define the tormenter let us quickly do
that you remember when I define the
atomic mass unit I told that one atomic
mass unit is 1/12 the mass of a carbon
atom which came out to be 1 point 6 7
into 10 to the power minus 27 kg s right
now how is it related to this mega
electron volt now remember Einsteins
mass-energy equivalence according to my
Einstein's mass energy equivalence this
mass has a mass energy associated with
it
what is that mass energy mass energy
associated with this mass will be e is
equal to MC square so this M we can put
it as this value that is 1 point 6 7
into 10 to the power minus 27 into C
square C is the speed of light that is 3
into 10 to the power 8 square so this
value comes out to be 1 point 4 9 to 4
into 10 to the power minus 10 joules so
I was able to convert this mass which
was earlier in kgs into joules because
kind strengthens the rule says that mass
is equivalent to energy so if I say it
has a mass of 1 point 6 7 into 10 to the
power minus 27 kg and if I say that it
has a mass energy of 1 point 4 9 4 into
10 to the power minus 10 joules
both mean the same thing because this
mass is equivalent to this mass energy
right so now our aim is to convert this
energy which is in joules into electron
volts so that is simpler for us right so
how do we convert it into electron volts
1 point 4 9 2 4 into 10 to the power
minus 10 divided by 1 point 6 into 10 to
the power minus 19 electron volts so
this comes out to be 931 point 5 mega
electron volts and this becomes a
relationship between atomic mass unit
and mega electron volts because mostly
in nuclear physics the unit which is
used for energy is mega electron volt I
mean hardly you will see atomic mass
unit or joules or electron volts
everywhere you would see this unit so
this is an important relationship
between atomic mass unit and mega
electron volts so one atomic mass unit
becomes 931 point 5 mega electron volt
but C square because here we have
multiplied it by C square so that means
this C square has been take has to be
taken into
come so this is divided by C square plus
K I have multiplied it with this C
square so this atomic mass unit is this
much okay so now we now here we got this
mass defect as 0.1 one two three six so
we have to calculate the binding energy
so how do I calculate the binding energy
this will be Delta n C square so this
Delta M zero point one one two three six
you can be converted into mega electron
volts so this becomes zero point 1 1 2 3
6 into 9 31 point 5 mega electron volt
per C square into C square like this I
have written 4 Delta n now this C Square
and C Square will get cancelled
therefore binding energy will come out
to be one zero four point six six three
three four mega electron volts so this
would be the binding energy of the
nitrogen nucleus okay so please make a
note here that this C Square from why
did we divided by C square that's
because the energy which we have
calculated here this energy comes out to
be 931 point 5 mega electron volt but we
wanted to convert this mass right so the
unit of mass would be energy by C square
right this energy was 931 point 5 mega
electron volt divided by C square so
that is why we write one atomic mass
unit is equal to 9 31 point 5 mega
electron volt by C square now let us
look at it thank you please visit exam
few calm to watch free educational
videos try 3 online tests get the best
quality study materials study from the
best tutors and mentors and much more
thank you once again