NEET-XII-Physics
46: The Nucleus
- #31A 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.Ans : Given:
Counts received per second = 50000 Counts/second
Number of active nuclei, N = 6 × 1016
Total counts radiated from the source, `` \frac{dN}{\,\mathrm{\,dt\,}}`` = Total surface area × 50000 counts/cm2
= 4 × 3.14 × 1 × 104 × 5 × 104
= 6.28 × 109 Counts
We know
`` \frac{dN}{dt}=\lambda N``
Here, λ = Disintegration constant
`` \therefore \lambda =\frac{6.28\times {10}^{9}}{6\times {10}^{16}}``
`` =1.0467\times {10}^{-7}``
`` =1.05\times {10}^{-7}{\,\mathrm{\,s\,}}^{-1}``
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