Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
Share

# 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 - Physics

#### 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.

#### Solution

Given:
Counts received per second = 50000 Counts/second
Number of active nuclei, N = 6 × 1016
Total counts radiated from the source, "dN"/"dt" = Total surface area × 50000 counts/cm2

= 4 × 3.14 × 1 × 104 × 5 × 104

= 6.28 × 109 Counts

We know

"dN"/"dt" = lambdaN

Here , lambda = Disintegration constant

therefore lambda = (6.28 xx 10^9)/(6 xx 10^16)

= 1.0467 xx 10^-7

= 1.05 xx 10^-7  "s"^-1

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution 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 Concept: Radioactivity - Alpha Decay.
S