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

Answer:

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

Explanation:

Radioactive decay follows first order kinetics.

Given:

Half life of Br-88 = 16.3 s

time = 35.0 s

$ln\frac{a_0}{a} =K\times t$

where,

$a_0$ = Initial concentration of radioactive substance

a = Amount left after time 't'

K = rate constant

Half-life = 0.693/K

$K = \frac{0.693}{16.3} = 0.0425 s^{-1}$

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

$ln\frac{a_0}{a} =K\times t$

$ln\frac{a_0}{a} =0.0425\times 35.0$

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

$ln\frac{100}{a} = 1.4875$

[a] = 22.6% = 23%

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