Question

Unreasonable Results

The manufacturer of a smoke alarm decides that the smallest current of $\alpha$ radiation he can detect is $1.00\mu A$.

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

Short answer

1. The activity in curies is$84.46\mathrm{C}\mathrm{i}$.
2. The source's necessary activity is really high.
3. The current is really strong.

Step-by-step solution

Concept Introduction

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

$\mathrm{R}=\frac{0.693\mathrm{N}}{{\mathrm{t}}_{1/2}}$

Where,

${\mathrm{t}}_{1/2}=$Half life

$\mathrm{R}=$Activity

$\mathrm{N}=$Number of atoms

Activity in curies

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

$\begin{array}{r}Q=2\left(1.6\times {10}^{-19}\mathrm{C}\right)\\ =3.2\times {10}^{-19}\mathrm{C}\end{array}$

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.

$\begin{array}{r}R=\frac{1\times {10}^{-6}\mathrm{A}}{3.2\times {10}^{-19}\mathrm{C}}\\ =3.125\times {10}^{12}\mathrm{B}\mathrm{q}\\ =84.46\mathrm{C}\mathrm{i}\end{array}$

Therefore, the activity in curies is $84.46Ci$.

Unreasonable about this result

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

Assumption of current

c) The current is really strong.