A radioactive nucleus decays by two different process. The half life of the first process is 5 minutes and that of the second process is 30 s. The effective half-life of the nucleus is calculated to be α/11 s. The value of α is __________.

Solution

Physics sir · 3 steps

Two decay routes run side by side, so add the decay constants, not the half-lives.

1. Rates add up. Both processes remove nuclei at the same time, so the total decay rate is the sum of the two. That means λ = λ₁ + λ₂. Since λ = (ln 2)/T, the ln 2 cancels and the half-lives add as reciprocals.

   1/T = 1/T₁ + 1/T₂

2. Same units first. T₁ = 5 min = 300 s and T₂ = 30 s. Then 1/T = 1/300 + 1/30 = (1 + 10)/300 = 11/300 s⁻¹.
3. So. T = 300/11 s. Matching this with α/11 s gives α = 300.

The common slip: 3630. Adding the half-lives directly gives T = 330 s and so α = 3630. For parallel decay the reciprocals add, not the half-lives.

The idea to keep. When a nucleus can decay by parallel routes, the decay constants add. The effective half-life is shorter than either one, just like resistors in parallel.

Answer from the NTA final answer key. Solution by PrepWiser.