# Half life period of 53 I 125 is 60 days. What percentage of the original radioactivity would be present after 180 days?

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In one half life, half (i.e., 50%) of the original content is left and the rest decays.

In 60 days, 50% of original content is left.

In 120 days (2 half lives), 25% of original content (50% of content left after 1 half live, i.e. 50% of 50%) is left.

and in 180 days (3 half lives), 12.5% of original content is left.

i.e. 1/8th of original quantity is left after 3 half lives.

Radioactive decay follows the following first-order law:

`A_t = A_0 *e^(-lambda*t)`

where At = activity at time t

Ao = initial activity

`lambda` = the decay constant

t = time elapsed

`ln ((A_t) / (A_0)) = -lambda*t`

`t= (-1/lambda)*ln ((A_t) / (A_0))`

`t_(1/2) = (-1/lambda)*ln (1/2)`

`t_(1/2) = (ln2) / lambda`

`t_(1/2) = 0.693/lambda`

( Half life is 60 days)

`:.lambda = 0.693/60`

Now activity after t = 180 days will be

`A_t = A_0 *e^(-lambda*180)`

`(A_t) /( A_0) = e^((-0.693*180)/60)`

`(A_t) / (A_0) = e-2.079`

`(A_t) / (A_0) = 0.125`

``Converting this into percentage the activity after 180 days will be 12.5%

Alternately we can also calculate by using the following equation

`A_t = (A_0 )/2^n` where n = number of half lives = time elapsed/half-life

Here n= 180/60 = 3

Therefore , `A_t = (A_0)/2^3`

`A_t = (A_0)/8=0.125*A_0`

Converting into percentage , the activity will be 12.5% of the original activity

Radioactive reactions follows 1st order kinetics.

Thus the governing equation for 1st order reactions is:-

k*t = ln{[A0]/[A]}

where, k = decay constant = ln2/(half life) = 0.693/60 = 0.0116 days^-1

t = time interval under consideration = 180 days

[A0] = initial concentration of the substance = 100%

[A] = concentration of the substance after time 't' = x

Now,

0.0116*180 = ln(100/x)

or, ln(100/x) = 2.079

or, 100/x = e^2.079

or, 100/x = 8

or, x = 12.5%