A point source emitting alpha particles is placed at a distance o… - Vidyadip

Question

A point source emitting alpha particles is placed at a distance of 1m from a counter which records any alpha particle falling on its 1cm² window. If the source contains 6.0 × 10¹⁶ active nuclei and the counter records a rate of 50000 counts/ second, find the decay constant. Assume that the source emits alpha particles uniformly in all directions and the alpha particles fall nearly normally on the window.

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Answer

Counts received per cm² = 50000 Counts/sec.
N = N₃o of active nucleic = 6 × 10¹⁶
Total counts radiated from the source = Total surface area × 50000 counts/cm²
= 4 × 3.14 × 1 × 10⁴ × 5 × 10⁴ = 6.28 × 10⁹ $\text{Counts}=\frac{\text{dN}}{\text{dt}}$
We know, $\frac{\text{dN}}{\text{dt}}=\lambda\text{N}$
$\lambda=\frac{6.28\times10^9}{6\times10^{16}}=1.0467\times10^{-7}=1.05\times10^{-7}\text{s}^{-1}$

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