Solved Examples on Rate of Decay Formula

Example 1: If U-238 has a half-life of 4.468 × 10⁹ years, determine its rate of decay constant.

Solution: 

The problem refers to half life of U-238, Half of the original sample has already decayed. Hence the ratio N₀/N_t = 0.5.

ln (N_t /N₀) = -λt

ln 0.5 = -λ × 4.468 × 10⁹

λ = 1.55 x 10^-10 years^-1

Rate of decay constant (λ) is 8.38 x 10^-11 years^-1

Example 2: If U-238 has a 35% life of 5.142 × 10⁹ years, determine its rate of decay constant.

Solution: 

35% half life of U-238 has already decayed, Hence the ratio N₀/N_t = 0.65 as 65 percent of original sample remains.

ln (N_t /N₀) = -λt

ln 0.65 = -λ × 5.142 × 10⁹

λ = 0.838 x 10^-10 years^-1

Rate of decay constant (λ) is 8.38 x 10^-11 years^-1

Example 3: Determine the amount of time it will take for 25% of a sample of U-238 to radioactively decay with a decay constant of 1.55 x 10^-10 years^-1.

Solution: 

Since 75% of the original sample is still present, the ratio N_t/N₀ = 0.75. Where 25% of the sample has undergone radioactive decay.

Rate of decay constant (λ) = 1.55 × 10^-10 years^-1 

ln (N_t /N₀) = -λt

ln  0.75 = -1.55 × 10^-10 years^-1 × t

t = 1.86 x 10⁹ years 

The amount of time for 25% radioactive U-238 decay is 1.86 x 10⁹ years 

Example 4: Determine the amount of time it will take for 45% of a sample of U-238 to radioactively decay with a decay constant of 1.55 x 10^-10 years^-1.

Solution: 

Since 55% of the original sample is still present, the ratio N_t/N₀ = 0.75. Where 45% of the sample has undergone radioactive decay.

Rate of decay constant (λ) = 1.55 × 10^-10 years^-1

ln (N_t /N₀) = -λt

ln  0.55 = -1.55 × 10^-10 years^-1 × t

t = 3.86 x 10⁹ years

The amount of time for 45% radioactive U-238 decay is 3.86 x 10⁹ years

Example 5: The half-life of PD-100 is 3.6 days. How many atoms will remain after 20.0 days, if one has 6.02 x 10²³ at the beginning?

Solution: 

Time = 20 days

Half-life = 3.6 days

Initial atoms = 6.02 ×10²³ atoms

Formula used to determine number of atoms after 20 days.

N = N₀ × 1/2 × t/t_1/2

N = 6.02 ×10²³ × 1/2 × 20/3.6 

    = 1.28 × 10²²

The number of atoms present is 1.28 × 10²² 

Radioactive decay is the release of alpha, beta, and gamma particles from unbalanced atoms known as radionuclides. Some substances, like uranium, have no stable forms and are therefore always radioactive. Radioactive substances are referred to as radionuclides. Rate of Decay is calculated to tell the exact amount of radioactive material that is being radiated. In this article, we will learn about the rate of decay formula and its examples in detail.