Q68PE. - College Physics (Urone)

Question

Unreasonable Results

The manufacturer of a smoke alarm decides that the smallest current of \({\rm{\alpha }}\) radiation he can detect is \(1.00\,\mu A\). 

1. Find the activity in curies of an \({\rm{\alpha }}\) emitter that produces a \(1.00\,\mu A\)current of \({\rm{\alpha }}\) particles.  
2. What is unreasonable about this result? 
3. What assumption is responsible?

Step-by-Step Solution

Answer

1. The activity in curies is \(84.46\,{\rm{Ci}}\).
2. The source's necessary activity is really high.
3. The current is really strong.

Concept Introduction

The following is the relationship between activity, half-life, and the number of atoms:

\({\rm{R = }}\frac{{{\rm{0}}{\rm{.693N}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\)

Where,

\({{\rm{t}}_{{\rm{1/2}}}}{\rm{ =  }}\)Half life

\({\rm{R  =  }}\)Activity

\({\rm{N  =  }}\)Number of atoms

Activity in curies

1.  

The nucleus of a helium atom is represented by alpha particles. As a result, an alpha particle's overall charge is

\(\begin{aligned}Q = 2\left( {1.6 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\\ = 3.2 \times {10^{ - 19}}\,{\rm{C}}\end{aligned}\)

The formula also relates the number of decays per unit time to current and charge:

\(R = \frac{I}{Q}\)

Where I denotes current and Q denotes charge

Now, in the preceding equation, enter in the values of I and Q and solve for the value of R.

\(\begin{aligned}R = \frac{{1 \times {{10}^{ - 6}}\;{\rm{A}}}}{{3.2 \times {{10}^{ - 19}}\,{\rm{C}}}}\\ = 3.125 \times {10^{12}}{\rm{\;Bq}}\\ = 84.46\,{\rm{Ci}}\end{aligned}\)

Therefore, the activity in curies is \(84.46\,Ci\).

Unreasonable about this result

b) The source's necessary activity is really high.

Assumption of current

c) The current is really strong.