Atumor is injected with 0.8 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?

Based on the radioactive decay model:

A = A₀exp(-kt) (1)

where

A = concentration of the radioisotope at time t

A₀ = initial concentration

k = decay constant

t = time

Now, based on the given data:

A₀ of I-125 = 0.8 g

Since the decay rate is 1.15% per day, the amount of I-125 left after t = 1 day is:   A = 0.8 - (1.15/100)*0.8 g = 0.7908 g

substituting for A, A₀ and t in eq(1) we get:

0.7908 = 0.8 exp(-k*1)

k = 0.01157 day-1

The decay constant is related to t1/2 as follows:

t1/2 = 0.693/k

= 0.693/0.01157 = 59.89 days

Ans: The t1/2 for I-125 approximately 60 days

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer