Q:

# 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?

Accepted Solution

A:
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 × $$(1-\dfrac{\textrm rate}{100})^{\textrm Time}$$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, $$(0.5)^{\dfrac{1}{x}}$$ = 0.9885Taking log both side log ( $$(0.5)^{\dfrac{1}{x}}$$ ) = Log 0.9885or, $$\dfrac{1}{x}$$ × log 0.5 = - 0.0050233or,  $$\dfrac{1}{x}$$ × ( - 0.301029 ) = - 0.0050233or, x =   $$\dfrac{0.301029}{0.0050233}$$ ∴ x = 59.92  ≈ 60 daysHence The Time period required for decay of Iodine-125 to half of its value is 60 days   . Answer