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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. Particles which can be added to the nucleus of an atom without changing its chemical properties are

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