Question

A certain radioactive isotope placed near a Geiger counter registers 160 counts per second. 32 hours later, the counter registers 10 counts per second. What is the half-life of the isotope?

A. 8 hours

B. 16 hours

C. 24 hours

D. 32 hours

E. none of the above

Answer 1

half life is 8.00 hours.

Step-by-step explanation:

The radioactive disintegration follows first order kinetics.

The integrated rate law for first order kinetics is:

`ln(([A_(0) ])/([A_(t)]))=kt`

Where

A0 = initial Geiger counter

At = final Geiger counter

t = time

k = rate constant

Let us calculate rate constant from the equation

`ln(([160])/([10]))=kX32`

k = 0.0866 / hour

the relation between half life and rate constant for first order reaction is:

k =0.693 / half life

half life =0.693 /k = 0.693 / 0.0866 = 8.00 hours

Answer 2

I believe that the answer for this question would be option A. 8 HOURS. Based on the given scenario above about a certain radioactive isotope placed near a Geiger counter, the half-life of the isotope 32 hours later would be 8 hours. Hope this is the answer that you are looking for.