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CBSE Class 12-science Answered

how derive a equation for average halflife of a radioactive substance?
Asked by Abhishika John | 13 Feb, 2018, 04:20: PM
Expert Answer
"Average half-life" is meaningless phrase because half-life is constant for given radioactive substance.
You can get the derivation for half-life from your physics text book or any physics book that has radioactivity as one of the chapter.
 
I assume that the student want 'mean-life' of radioactove substance and I am giving below the derivation of that

Radioactive decay is directly proportional to the amount of substance or number of atoms present at that  moment.
 
mathematically it means begin mathsize 12px style fraction numerator d N over denominator d t end fraction equals negative lambda N end style
where dN/dt is rate of disintegration, λ is decay constant and N is number of atoms present at that instant.
We can negelect the -ve sign in the above equation, it just indicates that number of atoms keep on decreasing.
Hence for dN atoms that lived between t and t+dt can be obtained from the above equation and it is given by

begin mathsize 12px style d N space equals space lambda cross times N cross times d t space equals space lambda cross times N subscript 0 e to the power of negative lambda t end exponent d t end style
In the above equation, i have used the radioactice decay equation begin mathsize 12px style N space equals space N subscript 0 space e to the power of negative lambda t end exponent end style
where N0 is the number of atoms present when t = 0
 
Since these dN atoms have individual life time between t and t+dt and dt is very small, we get lifetime for dN atoms as
 
begin mathsize 12px style d N cross times t space equals space lambda cross times N subscript 0 e to the power of negative lambda t end exponent space t space d t end style
since life time varies from 0 to infinity , we get total life time of N0 atoms as
begin mathsize 12px style integral d N cross times t space equals integral subscript 0 superscript infinity space lambda cross times N subscript 0 e to the power of negative lambda t end exponent space t space d t end style
mean life time begin mathsize 12px style tau space equals fraction numerator T o t a l space l i f e space t i m e space o f space N subscript 0 space a t o m s over denominator N subscript 0 end fraction space equals space lambda integral subscript 0 superscript infinity e to the power of negative lambda t end exponent space t space d t end style
above integration can be solved by considering begin mathsize 12px style integral u space d v space comma space w h e r e space u space equals space t space a n d space d v space equals space e to the power of negative lambda t end exponent d t end style
By solving the above integration we get maen life time begin mathsize 12px style tau space equals space 1 over lambda end style
 
 
Answered by Thiyagarajan K | 13 Feb, 2018, 05:48: PM

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