Question

A point source emitting alpha particles is placed at a distance of 1 m from a counter which records any alpha particle falling on its 1 cm2 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.

Answer 1

Explanation:

It is given:

Received counts per second = 50000 Counts/second

Active nuclei number, $N=6 \times 10^{16}$

From the source, the radiation total counts, dNdt = Total surface area × 50000 counts/cm2

$=4 \times 3.14 \times 5 \times 10^{4} \times 1 \times 10^{4}$

$=6.28 \times 10^{9} \text { Counts }$

It is known that dNdt = λN

where, λ is the constant of disintegration.

Therefore, $\lambda=6.28 \times 1096 \times 1016=1.05 \times 10-7 s-1$