Atumor 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?

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 ×

Or, 0.2 gm = 0.4 gm ×

or, =

Or, 0.5 =  

Or, = 0.9885

Taking log both side

log ( ) = Log 0.9885

or, × log 0.5 = - 0.0050233

or,   × ( - 0.301029 ) = - 0.0050233

or, x =  

∴ x = 59.92  ≈ 60 days

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

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25

22

step-by-step explanation: