Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains ${}^{10}\text{B}$ that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by ${}^{10}\text{B}$ nuclei and the formation of unstable form of ${}^{11}\text{B}$ undergoing alpha decay to ${}^7\text{Li}$ is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by ${}^{10}\text{B}$ and further alpha decay process.

Answer to Problem 19.69QP

Solution

The overall equation is given as,

${}_5^{10}\text{B}{+}_0^1\text{n}{\to }_3^7\text{Li}{+}_2^4\text{He}$

Explanation of Solution

Explanation

The equation representing the absorption of neutron by ${}^{10}\text{B}$ is given as,

${}_5^{10}\text{B}{+}_0^1\text{n}{\to }_5^{11}\text{B}$

The equation for the alpha decay in ${}^{11}\text{B}$ is given as,

${}_5^{11}\text{B}{\to }_3^7\text{Li}{+}_2^4\text{He}$

The overall equation is given as,

${}_5^{10}\text{B}{+}_0^1\text{n}{\to }_3^7\text{Li}{+}_2^4\text{He}$

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron $-10$ in the obtained reaction.

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron $-10$ is $\underset{\_}{-4.42\times 10^{-13}J}$.

Explanation of Solution

Explanation

Given

The mass of ${}^{10}\text{B}$ is $10.0129\text{amu}$.

The mass of ${}^1\text{n}$ is $1.00866\text{amu}$.

The mass of ${}^7\text{Li}$ is $7.01600\text{amu}$.

The mass of ${}^{\text{4}}\text{He}$ is $4.00260\text{amu}$.

The reaction is,

${}_5^{10}\text{B}{+}_0^1\text{n}{\to }_3^7\text{Li}{+}_2^4\text{He}$

The formula for the amount of energy released during fusion is given as,

$\Delta E=\Delta m{c}^2$ (1)

Where,

• $\Delta m$ is the change in the mass.
• $c$ is the speed of the light $\left(3.0\times {10}^8\text{m}/\text{s}\right)$.

The change in the mass is calculated as,

$\Delta m=\text{Mass of products}-\text{Mass of reactants}$

Substitute the mass of reactants and products in the above equation as,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\left(7.01600\text{amu}+\text{4}\text{.00260}\text{}\text{amu}\right)-\left(10.0129\text{amu}+1.00866\text{amu}\right)\\ =11.0186\text{amu}-11.02156\text{amu}\\ \text{=}-0.00296\text{amu}\end{array}$

The conversion of $\text{amu}$ into $\text{kg}$ is done as,

$1\text{amu}=1.66054\times {10}^{-27}\text{kg}$

Hence, the conversion of $-0.00296\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.00296\text{amu}=-0.00296\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.0049151984\times {10}^{-27}\text{kg}\end{array}$

Substitute the value of $\Delta m$ and $c$ in the equation (1),

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.0049151984\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-4.42\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

The conversion of $\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$1\text{kg}{\text{m}}^2{\text{s}}^{-2}=1\text{J}$

Hence, the conversion of $-4.42\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-4.42\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-4.42\times {10}^{-13}\times 1\text{J}\\ =\underset{\_}{-4.42\times 10^{-13}J}\end{array}$

Therefore, energy released by each nucleus of Boron $-10$ is $\underset{\_}{-4.42\times 10^{-13}J}$.

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

1. a. The overall equation is given as,

   ${}_5^{10}\text{B}{+}_0^1\text{n}{\to }_3^7\text{Li}{+}_2^4\text{He}$

2. b. The energy released by each nucleus of Boron $-10$ is $\underset{\_}{-4.42\times 10^{-13}J}$.
3. c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy