Question

you have 88g of a radioisotope. If the half-life of this radioisotope is 4 days, how many grams will remain after 12 days?

Answer 1

11g

Step-by-step explanation:

Answer 2

11.0 g.

Step-by-step explanation:

• The decay of radioisotope is a first order reaction.

• The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(4.0 days) = 0.1733 day⁻¹.

• The integration law of a first order reaction is:

kt = ln [A₀]/[A]

k is the rate constant = 0.1733 day⁻¹.

t is the time = 12.0 days.

[A₀] is the initial percentage of radioisotope = 88.0 g.

[A] is the remaining percentage of radioisotope = ??? g.

∵ kt = ln [Ao]/[A]

∴ (0.1733 day⁻¹)(12.0 days) = ln (88.0 g)/[A]

2.079 = ln (88.0 g)/[A]

• Taking exponential to both sides:

7.996 = (88.0 g)/[A]

∴ [A] = (88.0 g)/(7.996) = 11.0 g.