Question

World War II aircraft had instruments with glowing radium-painted dials (see figure ). The activity of one such instrument was $1.0\times {10}^5\mathrm{B}\mathrm{q}$ when new.

(a) What mass of ${}^{226}\mathrm{R}\mathrm{a}$ was present?

(b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument $57.0$years after it was made?

Short answer

a) The mass is $2.73\mu \mathrm{g}$.

b) The activity of this instrument is$9.76\times {10}^4\mathrm{B}\mathrm{q}$.

Step-by-step solution

Concept Introduction

For a given number of nuclei, the shorter the half-life, the more decays per unit time. As a result, activity R should be proportional to N, the number of radioactive nuclei, and inversely proportional to t1/2, their half-life. In reality, your instincts are spot on. It can be demonstrated that a source's activity is$R=\frac{0.693N}{t_{1/2}}$.

Find the mass

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

$\begin{array}{c}R=\frac{0.693N}{t_{1/2}}\\ N=\frac{R{t}_{1/2}}{0.693}\end{array}$

Where$t_{1/2}=$half-life ,$R=$activity and$N=$number of atoms.

And

$\begin{array}{c}{t}_{1/2}=\left(1.6\times {10}^3\mathrm{y}\right)\\ =5.05\times {10}^{10}\mathrm{s}\end{array}$

Now, in the previous equation, enter in the values of$\mathrm{R}$and$t_{1/2}$and solve for the value of N:

$\begin{array}{c}N=\frac{\left(1.0\times {10}^5\mathrm{B}\mathrm{q}\right)\left(5.05\times {10}^{10}\mathrm{s}\right)}{0.693}\\ =7.28\times {10}^{15}\\ N=7.28\times {10}^{15}\end{array}$

Now we must use the following formula to convert the number of atoms into mass:

$m=N\left(\frac{1\mathrm{m}\mathrm{o}\mathrm{l}}{6.02\times {10}^{23}}\right)\left(\frac{M}{1\mathrm{m}\mathrm{o}\mathrm{l}}\right)$

Now plugin the value of N and$226\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$for M and solve for the value of m

$\begin{array}{c}m=\left(7.28\times {10}^{15}\right)\left(\frac{1\mathrm{m}\mathrm{o}\mathrm{l}}{6.02\times {10}^{23}}\right)\left(\frac{226\mathrm{g}}{1\mathrm{m}\mathrm{o}\mathrm{l}}\right)\\ =2.73\times {10}^{-6}\mathrm{g}\\ =2.73\mu \mathrm{g}\\ m=2.73\mu \mathrm{g}\end{array}$

Therefore, the mass is$2.73\mu \mathrm{g}$.

Find the activity of this instrument

b) The following is the relationship between beginning activity and activity after time t:

$R=R_0{e}^{-\lambda t}$

In the above equation, plug in the values of$\lambda =\frac{0.693}{t_{1/2}},t=57\mathrm{y}$, and$t_{1/2}=1.6\times {10}^6\mathrm{y}$and solve to find the value of$\mathrm{R}$.

$\begin{array}{c}R=\left(1\times {10}^5\right)\exp \left(-\frac{0.693\times 57\mathrm{y}}{1.60\times {10}^3\mathrm{y}}\right)\\ =9.76\times {10}^4\mathrm{B}\mathrm{q}\\ R=9.76\times {10}^4\mathrm{B}\mathrm{q}\end{array}$

Therefore, the activity of this instrument is $9.76\times {10}^4\mathrm{B}\mathrm{q}$.