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The half-life of a radioactive isotope is 140d. How many days would it take for the decay rate of a sample of this isotope to fall to one-fourth of its initial value?

Short Answer

Expert verified

It would take 280 days for the decay rate of a sample of this isotope to fall to one-fourth of its initial value.

Step by step solution

01

The given data

  1. The half-life of the radioactive isotope,T1/2=140d
  2. The decay rate falls to one-fourth of its initial value.
02

Understanding the concept of decay rate

The radioactive decay is due to the loss of the elementary particles from an unstable nucleus to convert them into a more stable one. Thus, from the given formula for the rate of decay, consider the decay rate is inversely proportional to the half-life. Thus, using the condition that the rate of decay would fall to one-fourth, get the number of days of the substance for the new decay rate.

Consider the formula for the rate of decay as:

R=Nln2T1/2

03

Calculate the days for the decay rate to fall by one-fourth of its original value

By the definition of half-life and equation (i), the decay rate is reduced; then, the half-life of the substance has doubled. Thus, the same has reduced to12its initial amount after 140 d. So, reducing it to14=122 of its initial number requires that two half-lives have passed:

t=2T1/2=2140d=280d

Hence, the number of days is 280d.

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