Question

The no.of half lives elapsed before which 93.75% of a radioactive sample has decayed

Answer 1

Answer:

4 half lives

Explanation:

amount  of reactant left after n half life =(1/$2^{n}$)*(initial concentration)

let initial concentration be 100

then,

100-93.75=(1/$2^{n}$)*(100)

(1/$2^{n}$)=6.25/100

       =1/(100/6.25)

       =1/16

$2^{n}$ = 16

n=4

Hope it helps and please mark me as brainliest....

Answer 2

it is a first order reaction  after  this decay substance remaining will be 6.25

let consider initially there  was 100% substane

100⇒50 one half life

50⇒25 another half life

25⇒12.5     "

12.5⇒6.25   "

total 4 half lifes