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 :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

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.

$k=\frac{2.303}{11.46}\log\frac{200}{25}=0.1814$

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

Formula used : $t_{1/2}=\frac{0.693}{k}$

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

$t_{1/2}=\frac{0.693}{0.1814}=3.820$

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