L13 : Numericals - Nuclie, Physics, Class 12 Video Lecture - Class 12 | Best Video for Class 12

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the part 12 before going ahead with part
13 now let us try to solve some problems
based on what will be studied in this
section that is some problems related to
half life mean life and radioactive
decay so the first problem says a
radioactive isotope has a half-life of t
years how long will it take the activity
to reduce to 3.125%
of its original value so let us look at
it so it says that the half-life of the
sample is T years okay
what was the original value let us
assume that the original value was n not
let us also assume that after D K it
becomes n okay so now let us try to
first solve the first part of the
problem it says that it should how much
time it will take to for its activity to
reduce to 3.125% that means n should be
equal to 3.125%
of n not that that means n divided by n
naught should be equal to three point
one two five divided by hundred that is
equal to 1 by 32 so now let us make use
of the law of radioactive decay so from
the law of radioactive decay we know
that n is equal to n naught e to the
power minus lambda T or we can say n by
n naught is equal to e to the power
minus lambda T so from this we can say
we take natural log on both sides we can
say Ln n by n naught is equal to minus
lambda T so now we put this value of N
by n naught as 1 by 32 so when we put
this value
o of n by n knot we get Ln of 1 by 32 is
equal to minus lambda T or we can say Ln
1 minus Ln 32 is equal to minus lambda T
so Ln 1 is equal to 0 so this will be
equal to minus Ln 32 is equal to minus
lambda T so this minus minus will get
cancelled so from this we can say T is
equal to Ln 32 is three point four six
five seven divided by lambda now since
we know that half life that is T half is
given as 0.693 by lambda and we know
that T half in this question is given as
capital T so from this we can say that
lambda is equal to 0.693 divided by
capital T so we will put this value of
lambda in this equation so if we put
this value of lambda in this equation
then what do we get we get T is equal to
three point four six five seven divided
by zero point six nine three into T so
this comes out to be five T years so
that means after a time period of five
years the activity of the sample will
reduce to three point one two five
percentage of its original value now
similarly the second half is says that
the percentage should be one person that
means in the second half n divided by n
knot will be equal to one by 100 so that
is the only difference that will be
there otherwise everything will remain
the same so I leave the second part for
you to solve so that you can also have a
practice on how to solve such questions
so now let us look go ahead and look at
the next problem it says that obtain the
amount of cobalt-60 necessary to provide
a radioactive source of H mili Curie
strength the half-life of cobalt-60 is
five point three years so what are the
values given here half-life is given as
five point three years what else is
given here the strength is given here
what is the strength I mentioned before
also strength is nothing but the rate of
decay so strength will be nothing but DN
by DT because it is talking about the
radioactive source strength and which is
measured in Curie so it is basically
talking about the activity of the sample
so strength is equal to DN by DT that is
equal to eight million so we can convert
this H mili Curie into the SI unit that
is back here L so this will be 8 into
since it will also has milling so it
will be 10 to the power minus 3 into
Curie is 3 point 7 into 10 to the power
10 barrels first 10 megatons or you can
also write it as these many decays per
second right now let us make use of the
law of radioactivity the loss is that
the rate of decay is always proportional
to the initial number of nuclei in the
sample right so from this you can see
that because it is asking the amount
necessary so that means you will have to
calculate this n so this n will be equal
to 1 by lambda
DN by DT so this lambda can be written
as 0.693 divided by T half that is
half-life into DN by DT so now we can
put the values here so we get 0.693
divided by T half that is half-life
which will be five point three years so
you convert it into seconds so this will
be five point three into 365 because one
year has 365 days one day has 24 hours
one hour has 60 minutes and one minute
has 60 seconds so this much seconds so
this will be five point three into 365
into 24 into 6
stay in 260 this into DN by DT that is 8
into 10 to the power minus 3 into 3
point 7 into 10 to the power J so n
comes out to be 7 point 1 in 3 into 10
to the power 16 atoms so these many
atoms of cobalt-60 is required but it is
asking us to calculate the amount that
means you have to calculate the mass of
these many atoms of Cobalt
right now for Cobalt how many atoms are
present in 60 grams of hobart in 60
grams of cobalt you have 60 grams of
cobalt because the mass number is 60 so
the 60 grams of Kobal consists of 6.02 3
into 10 to the power 23 atoms this is
the Avogadro number that is the number
of atoms present in 1 mole of cobalt you
would have studied all these concepts in
your chemistry so 60 grams of cobalt
consists of 6.02 3 into 10 to the power
23 atoms so you have to calculate what
would be the mass of these many atoms of
cobalt so we can say that 6.02 3 into 10
to the power 23 atoms mass of these many
atoms is 60 grams
therefore mass of 7.13 3 into 10 to the
power 16 atoms would be how much it
would be 16 divided by 6 points 0 to 3
into 10 to the power 23 multiplied by 7
point 1 3 3 into 10 to the power 16 so
it comes out to be 7 point 1 0 6 into 10
to the power minus 6 grams so this would
be the amount of cobalt necessary to
provide this much of radioactive
strength right ok so now let us go ahead
and look at the third problem it says
that the half-life of strontium is 28
years so what is the disintegration rate
of 15
grams of this isotope so what do we have
we have T 1/2 as 28 years right so
convert it into second you get 28 into
365 into 24 into 60 into 16 seconds so
this would be the half-life now we have
to calculate the disintegration rate
that means the rate of decay so DN by DT
is to be calculated and the mass of this
isotope is given as 15 grams if you look
at this problem it is just the reverse
of whatever we did in problem number 2
so here the mass is given as 15
milligrams now for hope for the
strontium if you see the mass number is
19 so we can say that 90 grams of
strontium consists of how many atoms 6.0
2 3 into 10 to the power 23 atoms so how
many atoms will be present in 15 grams
of strontium it will is 6.02 3 into 10
to the power 23 divided by 19 into 15
milligrams so 15 into 10 to the power
minus 3 so it comes out to be 1 point 0
0 3 8 into 10 to the power 20 atoms so
we have to calculate that this
integration rate that means we have to
calculate DN by DT when we know that n
is equal to this much
so we have calculated n is equal to 1
point 0 0 3 8 into 10 to the power 20
atoms now from the law of radioactive
decay so from the law of decay we know
that DN by DT is equal to lambda n in
many other cases I have not mentioned
the negative sign because the negative
sign just shows that the number of
nuclei is decreasing with time so it has
nothing to do with the magnitude and
such so from this we can say DN by DT is
equal to lambda so what is lambda 0.693
divided by T 1/2 into n so let us put
the values now 0.6
9 3 into n that is 1 point 0 0 3 8 into
10 to the power 20 divided by T 1/2
which is 28 into 365 into 24 into 60
into 60 so this comes out to be 7 point
8 7 8 into 10 to the power 10 atoms per
second so this would be the
disintegration rate or the decay rate of
this sample so now I get solving these
problems would have given you some idea
on a half-life and average life and also
the law of radioactive decay and how to
apply them to solve problems so now let
us go ahead and start with the next
topic that is the types of radioactive
decay I mentioned I mentioned in one of
the previous slides that there are three
ways by which radioactivity can take
things or an unstable nuclei can DK by
three waves that is the alpha decay beta
decay and gamma decay so now I will
discuss each of these decays in detail
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