A point source emitting alpha particles is situated at a distance...

The total activity of the source is given by $A_0 = \lambda N$, where $\lambda$ is the decay constant and $N$ is the number of active nuclei.

Since the alpha particles are emitted uniformly in all directions, the fraction of particles hitting the window is the ratio of the area of the window to the total surface area of a sphere of radius $r$.

Fraction of particles registered $= \dfrac{A}{4\pi r^2}$

The count rate $C$ is the number of alpha particles striking the window per second:

$C = A_0 \times \dfrac{A}{4\pi r^2} = \lambda N \dfrac{A}{4\pi r^2}$

Rearranging for the decay constant $\lambda$:

$\lambda = \dfrac{C \times 4\pi r^2}{N \times A}$

Substituting the given values $C = 50000 \text{ s}^{-1}$, $r = 1 \text{ m} = 100 \text{ cm}$, $A = 1 \text{ cm}^2$, and $N = 6.0 \times 10^{16}$:

$\lambda = \dfrac{50000 \times 4\pi \times (100)^2}{6.0 \times 10^{16} \times 1}$

$\lambda = \dfrac{5 \times 10^4 \times 4\pi \times 10^4}{6.0 \times 10^{16}}$

$\lambda = \dfrac{20\pi \times 10^8}{6.0 \times 10^{16}}$

$\lambda = \dfrac{10\pi}{3} \times 10^{-8} \text{ s}^{-1}$

$\lambda \approx 10.47 \times 10^{-8} \text{ s}^{-1} = 1.047 \times 10^{-7} \text{ s}^{-1}$

Rounding to three significant figures, we get $1.05 \times 10^{-7} \text{ s}^{-1}$.