Question

nuclear decay occurs according to first-order kinetics. a nuclide decays in 3.40 days from 45.0 g to 12.1 g. what is the rate constant for the nuclide?

Answer 1

The rate constant for the nuclide according to first-order kinetics is 0.3863 per day.

Nuclear Decay in First-Order Kinetics:

m(t) = Me∧(-αt)

where m(t) = mass of nuclide after decay at time t;

M = mass of nuclide before decay;

α = rate constant of decay;

t = time for the decay;

What is first-order kinetics?

When a fixed percentage of the drug is excreted per unit of time, this is known as first-order kinetics.

For example, The amount of drug in the body has a direct relationship to the rate of removal. The amount of medication eliminated per unit of time increases with concentration. The medication concentration decreases by 50% for each half-life.

Now,

m(t) = Me∧(-αt)

⇒e∧(αt) = M/m(t)

applying ln on both sides;

αt = ln(m/m(t))

α = (ln(m/m(t))/t)

here t= 3.4 days, M = 45 grams, m(t) = 12.1 grams;

dropping these values in the above equation we get,

α = 0.3863 per day;

The rate constant for the nuclide according to first-order kinetics is 0.3863 per day.

To refer more about First Order Kinetics visit:

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