Question

A Geiger counter registers a count rate of 4,800 counts per minute from a sample of a radioisotope. Sixteen minutes later, the count rate is 1,200 counts per minute. What is the half-life of the radioisotope?

Answer 1

Answer : The half-life of the radioisotope is, 8.00 min

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

`k=(2.303)/(t)\log(a)/(a-x)`

where,

k = rate constant = ?

t = time passed by the sample = 16 min

a = initial amount of the reactant = 4800

a - x = amount left after decay process = 1200

Now put all the given values in above equation, we get

`k=(2.303)/(16)\log(4800)/(1200)`

`k=0.0866\text{ min}^(-1)`

Now we have to calculate the half-life.

`k=(0.693)/(t_(1/2))`

`0.0866\text{ min}^(-1)=(0.693)/(t_(1/2))`

`t_(1/2)=8.00\text{ min}`

Therefore, the half-life of the radioisotope is, 8.00 min