Chapter 20, Problem 82CP

(a)

Interpretation Introduction

Interpretation:

The number of $\alpha$ and $\beta$ particles when ${}^{238}U$ decays to ${}^{222}Rn$ is to be calculated.

Concept Introduction:

The decaying of particle can be called as spontaneous process in which one unstable subatomic particle transforms into multiple other particles. The particles which are created by this process will have less mass in comparison of the original particle.

Answer to Problem 82CP

The number of $\alpha$ particles = 4

The number of $\beta$ particles = 2

Explanation of Solution

The following equation is representing the decay of the uranium-232 to radon -222 −

  ${}_{92}^{238}U{\stackrel{\alpha }{\to }}_{90}^{234}Th{\stackrel{\beta }{\to }}_{91}^{234}Pa{\stackrel{\beta }{\to }}_{92}^{234}U{\stackrel{\alpha }{\to }}_{90}^{230}Th{\stackrel{\alpha }{\to }}_{88}^{226}Ra{\stackrel{\alpha }{\to }}_{86}^{222}Rn$

The following formula will be used to calculate the $\alpha$ - particles −

  $Numberofalphaparticle=\frac{differenceofmassnumber}{4}$

  $Numberofalphaparticle=\frac{238-222}{4}=4$

Atomic number of ${}^{238}U$ = 92

Atomic number of ${}^{222}Rn$ = 86

  $Numberofbetaparticles=2\times \left(numberofalphaparticleproduced\right)-\left(differenceinatomicnumbers\right)$

  $Numberofbetaparticles=2\times 4-\left(92-86\right)$

  $Numberofbetaparticles=2$

Hence, after decay four alpha particles and 2 beta particles will be produced.

(b)

Interpretation Introduction

Interpretation:

The reason of concern over the inhalation of ${}^{222}Rn$ is to be explained.

Concept Introduction:

The decaying of particle can be called as spontaneous process in which one unstable subatomic particle transforms into multiple other particles. The particles which are created by this process will have less mass in comparison of the original particle.

Answer to Problem 82CP

The inhalation or ingestion of ${}^{222}Rn$ will not cause any serious damage to the body because it is chemically inert and doesn’t stay in body for longer duration.

Explanation of Solution

The effectiveness of causing the damage to the body because of the inhalation or ingestion of ${}^{222}Rn$ is totally depends on the resistance time of radon. It is known that the radon is chemically inert, so it will pass out quickly through the body without causing any serious damage to the body.

(c)

Interpretation Introduction

Interpretation:

The identity of the solid and the balanced equation of the species decaying $\alpha$ particle are to be determined. The reason of solid being the more potent $\alpha$ −particle producer is to be explained.

Concept Introduction:

The decaying of particle can be called as spontaneous process in which one unstable subatomic particle transforms into multiple other particles. The particles which are created by this process will have less mass in comparison of the original particle.

Answer to Problem 82CP

The solid is Polonium.

The balanced equation of the species decaying $\alpha$ particle is given as −

  ${}_{86}^{222}Rn{\to }_{84}^{218}Po{+}_2^{4}He$

Explanation of Solution

The product of nuclide will be represented as ${}_Z^{A}X$ during the alpha decaying process of radon. Now, the chemical equation can be given as −

  ${}_{86}^{222}Rn{\to }_Z^{A}X{+}_2^{4}He$

The element ${}_Z^{A}X$ can be identified by the summation of A and Z which should be same on the both side of process. Hence, for the element X, the value of Z and A can be given as −

  $Z=86-2=84$

  $A=222-4=218$

So, the element ${}_Z^{A}X$ = ${}_{84}^{218}Po$

Now, the balanced equation will be given as −

  ${}_{86}^{222}Rn{\to }_{84}^{218}Po{+}_2^{4}He$

Therefore, the solid = Polonium. This element is known as high radioactive element which is more potent $\alpha$ −particle producer.

(d)

Interpretation Introduction

Interpretation:

The 4.0 phi per liter of air into concentration of unit ${}^{222}Rn$ atoms per liter of air and mole of ${}^{222}Rn$ per liter of air is to be converted.

Concept Introduction:

The decaying of particle can be called as spontaneous process in which one unstable subatomic particle transforms into multiple other particles. The particles which are created by this process will have less mass in comparison of the original particle.

Answer to Problem 82CP

The conversion of 4.0 phis per liter of air into concentration of unit ${}^{222}Rn$ atoms per liter of air is $7.0\times {10}^4\times \frac{{}^{222}Rnatoms}{Lair}$ .

The conversion of mole of ${}^{222}Rn$ per liter of air is $1.2\times {10}^{-19}\frac{mo{l}^{222}Rn}{Lair}$ .

Explanation of Solution

For ${}^{222}Rn$ = 4.0 phi per liter of air

The following conversion method will be used to convert the concentration unit of radon into atoms per liter of air −

  $1Ci=3.7\times {10}^{10}decayevents/s$

  $1pCi=1\times {10}^{-12}Ci$

The following formula will be used to calculate the decay constant using half −life −

  $\lambda =\frac{0.693}{t_{1/2}}$

  $\lambda =\frac{0.693}{3.82d}\times \frac{1d}{86400s}=2.10\times {10}^{-6}s$

Now, the following method will be used to calculate the value 4.0 phi per liter of air into concentration of unit ${}^{222}Rn$ atoms −

  $\frac{4.0pCi}{Lair}\times \frac{1\times {10}^{-12}Ci}{1pCi}\times \frac{3.7\times {10}^{10}atoms/s}{1Ci}\times \frac{1}{2.0\times {10}^{-6}{s}^{-1}}=7.0\times {10}^4\times \frac{{}^{222}Rnatoms}{Lair}$

Now, the following method will be used for the conversion of concentration units of moles of ${}^{222}Rn$ per liter of air −

  $7.0\times {10}^4\times \frac{{}^{222}Rnatoms}{Lair}\times \frac{1mol}{6.022\times {10}^{23}Rnatoms}=1.2\times {10}^{-19}\frac{mo{l}^{222}Rn}{Lair}$