Half-Life of Radioactive Sample Video Lecture | Crash Course for ENGAA

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12 so what do I mean by half-life of a
radioactive sample it means time at
which the number of nuclei reduces to
1/2 of their initial values that means
let us suppose you start with the say n
naught number of nuclei at time T is
equal to 0 so XT is equal to 0 you have
n naught number of nuclei so after now
with passage of time this number will
keep decreasing so after how much time
this number will become n naught by 2
that time is known as the half-life of
this nucleus so it is denoted by T 1/2
the subscript 1/2 denotes half-life now
it is basically a measure of how long
the radioactive nuclei masks because let
us suppose you have two different
samples one sample has a half-life of 1
year the second sample has a half-life
of two years that means the first sample
will become half of its initial value in
just one year the second sample will
become half of its initial value in two
years what is this defect this depicts
that the first sample will not last for
a longer time because in just one year
it became half right so again in another
1/2 here maybe it is no more be there so
the other sample it will become half in
two years so basically by looking at the
half-year we can get an idea of how long
that radio nuclei can last right so now
let us try to derive an expression for
half-life of a radioactive sample now we
know that from the law of radioactive
decay n is equal to n naught e to the
power minus lambda T so initially number
of sample is
and not now in order to determine
half-life later what would be the number
of samples it should be and not by 2 so
we can write n is equal to n naught by 2
here so I got by 2 is equal to n naught
e to the power minus lambda T so in
naught and n naught will get cancelled
so from this we can write e to the power
minus lambda T this T is nothing but T
1/2 okay
so T 1/2 is equal to 1 by 2 so from this
we can say minus lambda T half is equal
to Ln 1 by 2 because if you remove this
exponent if you take log on both sides
so you get this
so now minus lambda T 1/2 is equal to Ln
1 minus Ln 2 now Ln 1 is equal to 0 so
from this we can say lambda T half is
equal to Ln 2 so from this we can say T
half is equal to Ln 2 natural log of 2
is equal to 0.693 this divided by lambda
so this is the expression for half-life
of a radioactive sample so the half-life
of a radioactive sample depends upon the
decay constant now let us talk about the
mean life of a radioactive sample so
half-life was the time after which the
sample reduces to half of its initial
value
similarly mean life is nothing but the
average life of a nuclei in the
radioactive sample that means at an
average till and what would be the
lifespan of that radioactive sample
right now in this case any sample which
you have there so there will be so many
nuclei inside that sample so every
nuclei has an opportunity to decay any
nuclei candy-cane and any nuclei can
decay at any point of time right so how
do I calculate the average life of the
entire radioactive sample as a whole
Oh so let us have a look at this let us
suppose that n naught is the number of
radioactive sample at time T is equal to
zero now from the law of radioactive
decay we can see that DN by DT is equal
to minus lambda n right therefore the
number of nuclei which decay between
time period T and T plus DT will be what
is the number of nuclei that will decay
between this time DN that means these
many number of nuclei will decay in time
DT right okay so therefore number of
nuclei which decay between T and T plus
DT because in DT these many nuclei will
decay
so between time T and T plus DT how many
nuclei will decay that is lambda n deep
right now out of so these many nuclei
will decay between these time period
between T and T plus DT but now life of
some of these will be less whereas for
some of these it will be more for
example let us suppose if I tell you
that there are say say I take an example
of something let us take example of
human beings okay so if I say that
number of human beings who died between
2013 to 2020 is hundred I say the
hundred people die between 2013 to 2020
but we don't know how many people
exactly die at each year because some of
they might leave war some of them might
leave less right so it depends we don't
know exactly who will live till what
time so similar is the case here so here
you have this is the total number of
nuclei which will be K between T and T
plus DT but which nuclei will leave for
a longer time which nuclei will leave
for a lesser time that we are not sure
so we will have to calculate an average
life of this nuclear so how do we do
that so let us follow this approach so
this way I calculated the number of
nuclei which will decay between these
time period now let us try to calculate
the sum of the lives of all these nuclei
so what would be the sum of the lives of
all these nuclear if you if we assume
that the life of each nuclei is safety
so this will be lambda n DB into T so
this will be the sum of lives of all
these nuclei therefore if we integrate
this value from T is equal to 0 to T is
equal to infinity so that would become
the sum of all lives of all the nuclei
which constitute the radioactive sample
so using that we will be able to
calculate the average life of the sample
so let me denote the average life as T
average so this will be equal to
integration of T lambda n DT from zero
to infinity since I am calculating
average so this has to be divided by the
total number of sample so what was the
my total number of sample it was n
naught
okay so now we will integrate this so
now let us integrate this so this
becomes lambda into integration of T and
DT from zero to infinity divided by n
not now from the exponential law of
radioactive decay we know that n is
equal to n naught e to the power minus
lambda T so we put this value in this
expression so what do we get we get T
average is equal to lambda divided by n
not integration of T n naught e to the
power minus lambda T DT so this n naught
again being a constant we can take it
out so this becomes lambda n naught
divided by n naught integration of T e
to the power minus lambda T DT so this n
naught an N naught will get cancelled so
we get T average is equal to lambda into
integration of T e to the power minus
lambda T DT so we are integrating it
over T is equal to 0 to 2 is equal to
infinity so now we will be using the
product rule of integration in order to
solve this integration which you are
already away from your mathematics
classes so the product rule says that
integration of FX GX DX will be equal to
FX integration of GX DX minus
integration of integration of GX DX into
f dash X DX so this is the product rule
of integration in order to know further
details about it you can refer to the
mathematics video of class 12 so now
this will become equal to lambda t into
integration of e to the power minus
lambda T DT
- integration of integration e to the
power minus lambda T DT into F dash X
that is d by DT of T so DT by DT would
be 1 so this becomes 1 into DT so this
will become equal to T into integration
of e to the power minus lambda T this
will be equal to e to the power minus
lambda T divided by minus lambda so the
limits are 0 to infinity minus
integration of e to the power minus
lambda T by minus lambda into DT from 0
to infinity so this will be equal to
lambda into T now the value of t will be
infinity so e to the power infinity is 0
so anything multiplied by 0 would be 0
similarly when you put t is equal to 0
you again get 0 so the first term will
be 0 now for the next term integration
of e to the power minus lambda T DT
divided by minus lambda so this becomes
e to the power minus lambda T divided by
lambda square from 0 to infinity so this
becomes lambda by lambda square e to the
power infinity will again be 0 minus e
to the power 0 would be 1 so this will
be minus 1 so this lambda and this
lambda will cancel out so you get this
as 1 by lambda so what do we get we see
that the average life of a radioactive
sample comes out to be 1 by lambda so
one interesting thing which we observe
that both the half-life and average life
they are both dependent on the decay
constant for for radioactive
DK so from these two relations we can
derive a relationship between average
life and half life so T averages one by
lambda and what was T half T half life
was 0.693 by lambda right so using these
two expressions what can be saved
so from these two expressions we can say
that T average is equal to we can write
1 by lambda as 0.693 divided by T 1/2
right so we can write it as T average is
equal to T 1/2 divided by 0.693 so this
is the relationship between average life
and half-life of a radioactive sample so
basically this half-life and average
life they give us an idea about how long
or the lifespan of a radioactive nuclei
so normally the radioactive substances
which have a very small half-life are
generally not seen as as such in nature
for example cutonium is one such
radioactive substance it has such a hot
shot half-life that it decays very
quickly so it is not very easily
available in nature thank you please
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