Q. 125.0( 1 Vote )

Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half – life is 1590 years. What percentage will disappear in one year?


Answer :

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

According to 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 its half life = 1590 years,


From equation(3),we have


– kt = log (


– k×1590 = log (


– k×1590 =


– k×1590 = – log 2


k =


The equation becomes


log ( = – t


log ( = – 0.9996 t


Percentage Disappeared = (1 – 0.9996) × 100 = 0.04 %


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