Chapter 20, Problem 3SEPP

Interpretation Introduction

Interpretation:

The power output generated at different times due to the decay of ${}^{\text{238}}\text{Pu}$ is to be calculated.

Concept introduction:

The amount of a particular radioactive isotope left after time t is given as:

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{o}}}\text{=}-kt$

Here, $k$

is the rate constant for the radioactive decay, ${\text{N}}_{\text{t}}$ is the amount of sample left after time t, and ${\text{N}}_{\text{0}}$ is the initial amount of the sample.

Nearly all radioactive decays are of first order and the rate constant is given as:

$k\text{ = }\frac{0.693}{t_{1/2}}$

Here, $t_{1/2}$

is the half-life of the radioactive substance.

The rate of decay for a particular isotope is given by the rate law as:

$\text{Rate = }k\text{N}$

Here, $k$

is the rate constant for the radioactive decay and $\text{N}$ is the number of atoms in the sample. Rate is also called as activity of the isotope.

Answer to Problem 3SEPP

Correct answer: Option (d).

Explanation of Solution

Given information: Initial mass of ${}^{\text{238}}\text{Pu}$

$=\text{1mg}$

Energy of the alpha particle per decay $=\text{9}\text{.0}\times {\text{10}}^{-\text{13}}$

J

At first, calculate the number of ${}^{\text{238}}\text{Pu}$ atoms present in $\text{1mg}$

of sample.

It is known that one mol of any substance is equivalent to its molar mass.

Thus, the conversion factor for ${}^{238}\text{Pu}$ is $\left(\frac{1\text{ mol}}{238{\text{ g }}^{\text{238}}\text{Pu}}\right)$.

Moreover, it is also known that one mol of any substance is equivalent to Avogadro’s number.

Thus, the conversion factor for ${}^{238}\text{Pu}$ is $\left(\frac{6.022\times {\text{ 10}}^{\text{23}}\text{ atoms}}{1\text{ mol}}\right)$

Also, one gram is equivalent to $1000$ milligrams (mg).

Thus, the conversion factor is $\left(\frac{1\text{ g}}{1000\text{ mg}}\right)$.

Hence, by using the conversion factor, the number of atoms of ${}^{238}\text{Pu}$

present in $1\text{ mg}$ is calculated as follows:

$\left(1.00{\text{ mg }}^{\text{238}}\text{Pu}\right)\left(\frac{1\text{g}}{1000\text{ mg}}\right)\left(\frac{1\text{ mol}}{238{\text{ g }}^{\text{238}}\text{Pu}}\right)\left(\frac{6.022\times {\text{10}}^{\text{23}}\text{atoms}}{1\text{ mol}}\right)=\text{2}\text{.530}\times {\text{10}}^{18}{\text{atoms}}^{\text{238}}\text{Pu}$

Thus, the initial number of atoms in one milligram of ${}^{238}\text{Pu }$

is $\text{2}\text{.530 }\times {\text{ 10}}^{18}\text{ atoms }$

Now, from the result of SEPP question 2, it is clear that one atom of ${}^{238}\text{Pu}$

decays to give one alpha particle.

Hence, $\text{2}\text{.530 }\times {\text{ 10}}^{18}\text{ atoms }$

of ${}^{238}\text{Pu}$

decay to give $\text{2}\text{.530 }\times {\text{ 10}}^{18}$ alpha particle.

Now, the activity of ${}^{238}\text{Pu }$ is calculated as follows:

$\text{Activity = }k\text{N}$

Substitute the value of rate constant from the result of SEPP question 1 to get:

$\begin{array}{c}\text{Activity}=\text{(2}\text{.6}\times {\text{10}}^{-\text{10}}{\sec }^{-1}\text{)}\times \text{(2}\text{.530}\times {\text{10}}^{\text{18}}\text{ atoms)}\\ \text{Activity}=\text{6}\text{.578}\times {\text{10}}^8\text{ decays/sec }\end{array}$

The energy of the alpha particles is as follows:

$\begin{array}{c}\text{Energy }=\text{(6}\text{.578}\times {\text{10}}^8\text{decay}{\sec }^{-1}\text{)}\times \text{(9}\times {\text{10}}^{-13}\text{ J/decay)}\\ \text{Energy }\approx \text{ 58}\times {\text{10}}^{-5}\text{ J/sec }\\ \text{= 58}\times {\text{10}}^{-5}\text{ W }\end{array}$

It is known that one watt is equivalent to $1000$ milliwatts.

Thus, the conversion factor is $\left(\frac{1000\text{ mW}}{1\text{ W}}\right)$.

The conversion of energy is as follows:

$\begin{array}{c}\text{Energy }=\text{58 }\times {\text{ 10}}^{-5}\cancel{\text{W}}\times \left(\frac{1000\text{ mW}}{1\cancel{\text{W}}}\right)\\ =\text{0}\text{.58 mW}\end{array}$

Thus, the power dissipated at $t=0$ is $\text{0}\text{.58 mW}$.

Similarly, the power dissipated at $t=10\text{ years}$ is calculated as:

At first, convert the given time into seconds as follows:

The relation between days and year is represented below.

$1\text{ year = 365 days}$

Thus, the conversion factor is $\frac{365\text{ days}}{1\text{ year}}$

Similarly,

$1\text{ day = 24 hours}$

Thus, the conversion factor is $\frac{\text{24 hour}}{1\text{ day}}$

Also,

$1\text{ hour = 3600 seconds}$

Thus, the conversion factor is $\frac{\text{3600 seconds}}{1\text{ hour}}$.

Now, convert the given time in seconds as follows:

$\begin{array}{c}\text{time = }\left(\text{10 }\cancel{\text{years}}\times \frac{365\cancel{\text{days}}}{1\cancel{\text{year}}}\times \frac{24\cancel{\text{hours}}}{1\cancel{\text{day}}}\times \frac{3600\text{ seconds}}{1\cancel{\text{hour}}}\right)\\ =3.15\times {\text{ 10}}^8\text{sec}\end{array}$

The number of atoms of ${}^{238}\text{Pu }$

decaying in ten years can be calculated as follows:

$\text{ln}\frac{N_t}{N_o}\text{=}-kt$

Here, $N_o$ is initial number of atoms while $N_t$ is the final number of atoms.

Substitute the values of $k$, $t$, and $N_o$ in the above expression,

$\begin{array}{c}\text{ln}\frac{N_t}{2.530\times {\text{ 10}}^{\text{18}}\text{atoms}}=-\text{ (2}\text{.6 }\times {\text{ 10}}^{-\text{10}}{\text{ sec}}^{-\text{1}})\times \text{ (3}\text{.15 }\times {\text{ 10}}^{\text{8}}\text{ sec)}\\ \text{ln}\frac{N_t}{2.530\times {\text{ 10}}^{\text{18}}\text{atoms}}=-\text{ (0}\text{.0819})\\ \frac{N_t}{2.530\times {\text{ 10}}^{\text{18}}\text{atoms}}=e^{-\text{ (0}\text{.0819})}\\ N_t=\text{2}\text{.303 }\times {\text{ 10}}^{\text{18}}{\text{ atoms }}^{238}\text{Pu }\end{array}$

The activity of ${}^{238}\text{Pu }$ is given as follows:

$\text{Activity = }k\text{N}$

Substitute $\text{2}\text{.6}\times {\text{10}}^{-\text{10}}{\sec }^{-1}$ for $k$ and $\text{2}\text{.330}\times {\text{10}}^{\text{18}}\text{ atoms}$ for N in the above expression.

$\begin{array}{l}\text{Activity}=\text{(2}\text{.6 }\times {\text{10}}^{-\text{10}}{\sec }^{-1}\text{)}\times \text{(2}\text{.330 }\times {\text{10}}^{\text{18}}\text{ atoms)}\\ \text{Activity}=\text{6}\text{.060 }\times {\text{10}}^8\text{ decays/sec }\end{array}$

The energy of the alpha particles is calculated as follows:

$\begin{array}{c}\text{Energy }=\text{ (6}\text{.060 }\times {\text{10}}^8\text{decay}{\sec }^{-1}\text{)}\times \text{(9 }\times {\text{ 10}}^{-13}\text{ J/decay)}\\ \text{Energy }\approx \text{ 53}\times {\text{10}}^{-5}\text{ J/sec }\\ \text{= 53}\times {\text{10}}^{-5}\text{ W }\end{array}$

It is known that one watt is equivalent to $1000$

milliwatts.

Thus, the conversion factor is $\left(\frac{1000\text{ mW}}{1\text{ W}}\right)$.

Energy is converted as follows:

$\begin{array}{c}\text{Energy }=\left(\text{53 }\times {\text{ 10}}^{-5}\cancel{\text{W}}\right)\times \left(\frac{1000\text{ mW}}{1\cancel{\text{W}}}\right)\\ =\text{0}\text{.53 mW}\end{array}$

Thus, the power dissipated at $t=10\text{ years}$ is $\text{0}\text{.53 mW}$.

Hence, option (d) is correct.

Reasons for the incorrect option:

Option (a) is incorrect because the power dissipated at $t=0$ and at $t=10\text{ years}$ are not equal to $0.081\text{ mW and 0}\text{.075 mW }$, respectively.

Option (b) is incorrect because the power dissipated at $t=0$ and at $t=10\text{ years}$ are not equal to $\text{2}\text{.6}\times {\text{10}}^{-\text{10}}\text{mW and 2}\text{.4 }\times {\text{10}}^{-\text{10}}\text{mW }$, respectively.

Option (c) is incorrect because the power dissipated at $t=0$ and at $t=10\text{ years}$ are not equal to $\text{1 mW and 0}\text{.92 mW }$, respectively.

Hence, options (a), (b), and (c) are incorrect.