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
[tex]ln\frac{a_0}{a} =K\times t[/tex]
where,
[tex]a_0[/tex] = Initial concentration of radioactive substance
a = Amount left after time 't'
K = rate constant
Half-life = 0.693/K
[tex]K = \frac{0.693}{16.3} = 0.0425 s^{-1}[/tex]
now, substitute the value of rate constant and t (35.0 s) in the formula,
[tex]ln\frac{a_0}{a} =K\times t[/tex]
[tex]ln\frac{a_0}{a} =0.0425\times 35.0[/tex]
Let a0 (initial concentration of bromine-88) be 100.
[tex]ln\frac{100}{a} = 1.4875[/tex]
[a] = 22.6% = 23%
Percentage of bromine-88 left after 35.0 s = 23%
