Answer :

Let the quantity of radium at any time t be A.

According to the question,



where k is a constant




Integrating both sides, we have


= – k∫dt


log|A| = – kt + c……(1)


Given, Initial quantity of radium be A0 when t = 0 sec


Putting the value in equation (1)


log|A| = – kt + c


log| A0| = 0 + c


c = log| A0| ……(2)


Putting the value of c in equation (1) we have,


log|A| = – kt + log| A0|


log|A| – log| A0| = – k t []


log ( = – kt ……(3)


Given that the radium decomposes 1.1% in 25 years,


A = (100 – 1.1)% = 98.9% = 0.989 A0 at t = 25 years


From equation(3),we have


– kt = log (


– k×25 = log (


k = –


The equation becomes


log ( = – t


Now,


log ( = – t


log ( = – t


= – t


(log 2 = 0.6931 and log 0.989 = 0.01106)




t = 1567 years


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