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 $1cm^2$ window. If the source contains $6.0 \times 10^{16}$ 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 $cm^2 = 50000$ Counts/sec.
$N = N_3o$ of active nucleic $= 6 \times 10^{16}$
Total counts radiated from the source = Total surface area $\times 50000$ counts/cm$^2$
$= 4 \times 3.14 \times 1 \times 10^4 \times 5 \times 10^4 = 6.28 \times 10^9$
$\text{Counts}=\frac{\text{dN}}{\text{dt}}$
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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