IDNLearn.com makes it easy to get reliable answers from experts and enthusiasts alike. Explore thousands of verified answers from experts and find the solutions you need, no matter the topic.

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


Sagot :

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 : 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 :

[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]

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.

[tex]k=\frac{2.303}{11.46}\log\frac{200}{25}=0.1814[/tex]

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

Formula used : [tex]t_{1/2}=\frac{0.693}{k}[/tex]

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

[tex]t_{1/2}=\frac{0.693}{0.1814}=3.820[/tex]

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

We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.