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Question

The half life of a radioactive substance is 10 years and its initial mass is 1 g. The remaining amount after 20 years is ________.

The correct answer is

0.25 g

Radioactive Substance Half-Life: Understanding Decay

The concept of half-life is fundamental in understanding the decay of radioactive substances. It represents the time required for half of the radioactive nuclei in a sample to undergo decay.

Half-Life Definition

  • The half-life ($\(T_{1/2}\)$) of a radioactive substance is a constant value that indicates how quickly it decays.
  • After one half-life, the amount of the original radioactive substance reduces to 50%.
  • After two half-lives, it reduces to 25%, and so on. This is an exponential decay process.

Calculating Number of Half-Lives

To determine the remaining amount of a radioactive substance after a certain period, we first need to find out how many half-lives have passed during that time. The question states that the half-life ($\(T_{1/2}\)$) is 10 years and the total time ($\(t\)$) elapsed is 20 years.

The number of half-lives ($\(n\)$) is calculated using the formula:

\[n = \frac{\text{Total time }(t)}{\text{Half-life }(T_{1/2})}\]

Substituting the given values:

\[n = \frac{20 \text{ years}}{10 \text{ years}} = 2 \text{ half-lives}\]

This means that the substance has undergone decay through two half-life periods.

Remaining Mass of Radioactive Substance

The initial mass of the radioactive substance is given as 1 g. We need to find the remaining amount after 2 half-lives. The general formula to calculate the remaining amount ($\(N\)$) after 'n' half-lives from an initial amount ($\(N_0\)$) is:

\[N = N_0 \left(\frac{1}{2}\right)^n\]

Here, \(N_0 = 1 \text{ g}\) and \(n = 2\).

Substituting these values into the formula:

\[N = 1 \text{ g} \times \left(\frac{1}{2}\right)^2\]

\[N = 1 \text{ g} \times \left(\frac{1}{4}\right)\]

\[N = 0.25 \text{ g}\]

Alternatively, we can track the decay step-by-step:

Time Elapsed Number of Half-Lives Remaining Mass
Initial (0 years) 0 1 g
After 10 years 1 $\(1 \text{ g} \times \frac{1}{2} = 0.5 \text{ g}\)$
After 20 years 2 $\(0.5 \text{ g} \times \frac{1}{2} = 0.25 \text{ g}\)$

Therefore, after 20 years, the remaining amount of the radioactive substance is 0.25 g.

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Important Questions from Radioactivity

  1. The energies of the 3 lowest states of an atom are E 0 = −14 eV, E 1  = −9 eV and E 2  = −7 eV. The Einstein coefficients are A 10  = 3 × 10 8  s −1 , A 20  = 1.2 × 10 8  s −1  and A 21  = 8 × 10 7  s −1 . If a large number of atoms are in the energy level E 2 , the mean radiative lifetime of this excited state is
  2. The nuclei of 137 Cs decay by the emission of β - particles with a half life of 30.08 years. The activity (in units of disintegrations per second or Bq) of a 1 mg source of 137 Cs, prepared on January 1, 1980, as measured on January 1, 2021 is closest to

  3. The Q - value of the α - decay of 232 Th to the ground state of 228 Ra is 4082 keV. The maximum possible kinetic energy of the α - particle is closest to

  4. Radioactivity is the characteristic of which of the following?

  5. A free neutron decays spontaneously into a proton, an electron and ___________ .

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