Question

What is the half-life of a radioisotope if 25.0 grams of an original 200.-gram sample of the isotope remains unchanged after 11.46 days?

(1) 2.87 d (3) 11.46 d
(2) 3.82 d (4) 34.38 d

Answer 1

1. 200/2=100. 100/2=50. 50/2=25. So that's 3 to get to 25.

2. 11.46/3=3.82

The answer is (2).

Answer 2

Answer : The correct option is, (2) 3.82 d

Solution : Given,

As we know that the radioactive decays follow first order kinetics.

So, the 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 taken for decay process = 11.46 days

a = initial amount of the reactant = 200 g

a - x = amount left after decay process = 25 g

Putting values in above equation, we get the value of rate constant.

`k=(2.303)/(11.46)\log(200)/(25)=0.1814`

Now we have to calculate the half life of a radioisotope.

Formula used :
`t_(1/2)=(0.693)/(k)`

Putting value of 'k' in this formula, we get the half life.

`t_(1/2)=(0.693)/(0.1814)=3.820`

Therefore, the half-life of a radioisotope is, 3.820 d