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: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 gramThe rate of decay = 1.15 %Let The time period for decay = x day The finial quantity after decay = half of initial quantityI.e The finial quantity after decay = 0.2 gramNow ,The final quantity after decay = initial quantity × [tex](1-\dfrac{\textrm rate}{100})^{\textrm Time}[/tex]Or, 0.2 gm = 0.4 gm × [tex](1-\dfrac{\textrm 1.15}{100})^{\textrm x}[/tex]or, [tex]\frac{0.2}{0.4}[/tex] = [tex](0.9885)^{x}[/tex]Or, 0.5 =  [tex](0.9885)^{x}[/tex]Or, [tex](0.5)^{\dfrac{1}{x}}[/tex] = 0.9885Taking log both side log ( [tex](0.5)^{\dfrac{1}{x}}[/tex] ) = Log 0.9885or, [tex]\dfrac{1}{x}[/tex] × log 0.5 = - 0.0050233or,  [tex]\dfrac{1}{x}[/tex] × ( - 0.301029 ) = - 0.0050233or, x =   [tex]\dfrac{0.301029}{0.0050233}[/tex] ∴ x = 59.92  ≈ 60 daysHence The Time period required for decay of Iodine-125 to half of its value is 60 days   . Answer