The count rate of nuclear radiation emitted from a radioactive...

The radioactive decay equation is given by:

$R = R_0 e^{-\lambda t}$

Taking the natural logarithm on both sides:

$\ln R = \ln R_0 - \lambda t$

This represents the equation of a straight line $y = mx + c$, where $y = \ln R$, $x = t$, and the slope is $m = -\lambda$.

Using the data points at $t_1 = 25 \text{ min}$ and $t_2 = 100 \text{ min}$:

$R_1 = 16 \times 10^9 \text{ s}^{-1}$

$R_2 = 2.0 \times 10^9 \text{ s}^{-1}$

The slope $m$ is given by:

$m = \dfrac{\ln R_2 - \ln R_1}{t_2 - t_1} = \dfrac{\ln(2.0) - \ln(16)}{100 - 25}$

$m = \dfrac{\ln\left(\dfrac{2.0}{16}\right)}{75} = \dfrac{\ln\left(\dfrac{1}{8}\right)}{75} = \dfrac{-3 \ln 2}{75} = \dfrac{-\ln 2}{25}$

Since $\lambda = -m$:

$\lambda = \dfrac{\ln 2}{25} \approx \dfrac{0.693}{25} = 0.0277 \text{ min}^{-1} \approx 0.028 \text{ min}^{-1}$

This is consistent with the other data points, confirming the decay constant is approximately $0.028 \text{ min}^{-1}$.