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? 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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