Question

Bromine-88 is radioactive and has a half life of 16.3 seconds. What percentage of a sample would be left after 35.0 seconds? Round your answer to 2 significant digits .

Answer 1

Percentage of bromine-88 left after 35.0 s = 23%

Step-by-step explanation:

Radioactive decay follows first order kinetics.

Given:

Half life of Br-88 = 16.3 s

time = 35.0 s

`ln(a_0)/(a) =K* t`

where,

`a_0` = Initial concentration of radioactive substance

a = Amount left after time 't'

K = rate constant

Half-life = 0.693/K

`K = (0.693)/(16.3) = 0.0425 s^(-1)`

now, substitute the value of rate constant and t (35.0 s) in the formula,

`ln(a_0)/(a) =K* t`

`ln(a_0)/(a) =0.0425* 35.0`

Let a0 (initial concentration of bromine-88) be 100.

`ln(100)/(a) = 1.4875`

[a] = 22.6% = 23%

Percentage of bromine-88 left after 35.0 s = 23%