If 'f' denotes the ratio of the number of nuclei decayed (N d ) to the number of nuclei at t = 0 (N₀) then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as :
[λ is the radioactive decay constant]

Solution

Physics sir · 3 steps

Decayed nuclei only ever increase, so f grows with time and its slope has to be positive.

1. Nuclei left. By the decay law, the number still undecayed at time t is N = N₀ e^(-λ t).
2. Nuclei decayed. Every nucleus that is gone has decayed, so N_d = N₀ - N = N₀(1 - e^(-λ t)). Dividing by N₀ gives f = 1 - e^(-λ t).

   df/dt = d/dt(1 - e^(-λ t)) = λ e^(-λ t)

3. So. The rate of change of f is λ e^(-λ t), which is option (c). It is positive and shrinks over time, as it should: decays keep adding up, but more slowly as fewer nuclei remain.

The idea to keep. Always ask which quantity you are tracking. Nuclei left fall as e^(-λ t); nuclei decayed rise as 1 - e^(-λ t). The decayed fraction always has a positive rate.

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