Question

A tumor is injected with 0.4 grams of Iodine-125, which has a decay rate of 1.15% per day. To the nearest day, how long will it take for half of the Iodine-125 to decay?

Answer 1

Answer:

The Time period required for decay of Iodine-125 to half of its value is 60 days   .

Step-by-step explanation:

Given as :

The initial quantity of iodine-125 = 0.4 gram

The rate of decay = 1.15 %

Let The time period for decay = x day

The finial quantity after decay = half of initial quantity

I.e The finial quantity after decay = 0.2 gram

Now ,

The final quantity after decay = initial quantity × $(1-\dfrac{\textrm rate}{100})^{\textrm Time}$

Or, 0.2 gm = 0.4 gm × $(1-\dfrac{\textrm 1.15}{100})^{\textrm x}$

or, $\frac{0.2}{0.4}$ = $(0.9885)^{x}$

Or, 0.5 =  $(0.9885)^{x}$

Or, $(0.5)^{\dfrac{1}{x}}$ = 0.9885

Taking log both side

log ( $(0.5)^{\dfrac{1}{x}}$ ) = Log 0.9885

or, $\dfrac{1}{x}$ × log 0.5 = - 0.0050233

or,  $\dfrac{1}{x}$ × ( - 0.301029 ) = - 0.0050233

or, x =   $\dfrac{0.301029}{0.0050233}$

∴ x = 59.92  ≈ 60 days

Hence The Time period required for decay of Iodine-125 to half of its value is 60 days   . Answer