The activity of a radioactive substance, having a decay constant $$\lambda$$ and number of nuclei N at time t, is given by

$$A = \left| {dN/dt} \right| = \lambda N = \lambda {N_0}{e^{ - \lambda t}}$$ ..... (1)

Take logarithm on both sides of equation (1) to get

$$\ln \left| {dN/dt} \right| = \ln (\lambda {N_0}) - \lambda t$$ ...... (2)

Thus, the graph between t and $$\left| {dN/dt} \right|$$ is a straight line with slope $$ - \lambda $$.

Slope $$ = - \lambda = {{3 - 4} \over {6 - 4}}$$ (From graph) or $$\lambda = {1 \over 2}$$ year^$$-$$1

Half life $${T_{1/2}} = {{0.693} \over \lambda } = 2 \times 0.693$$ years = 1.386 years

4.16 years is approximately 3 half-lives

Nuclei will decay by a factor of 2³ = 8

$$\therefore$$ p = 8