Chapter 21, Problem 21.47QA

Interpretation Introduction

To find:

a) The balanced nuclear equation for decay of $C$.

b) The binding energy of $C$..

Answer to Problem 21.47QA

Solution:

a) Balanced nuclear equation for decay of $C$ is

$\mathrm{C}\to \mathrm{B}+\mathrm{{\rm B}}$

b) The binding energy of $C$ is $1.176 \times {10}^{-11}J$.

Explanation of Solution

1) Concept:

First, we will balance the nuclear equation for decay of $C$ by using the emission of beta particles. To determine the binding energy of $C$, we need to calculate the mass decay of $C$. From the mass decay and Albert Einstein equation, we will calculate the binding energy.

2) Formula:

Albert Einstein equation for the binding energy.

$\mathrm{B}\mathrm{E}=(∆\mathrm{m}){\mathrm{c}}^2$

3) Given:

i) The exact mass of $C$ is $1.82850\times {10}^{-26} kg$.

ii) $1 J = 1 kg {\left(\frac{m}{s}\right)}^2$

iii) $$ Mass of proton = $1.67262\times {10}^{-27}\mathrm{k}\mathrm{g}$

iv)  Mass of neutrons = $1.67493\times {10}^{-27}\mathrm{k}\mathrm{g}$

v) Mass of electrons = $9.10939\times {10}^{-31}kg$

vi) Speed of light c = $2.998 \times {10}^{8}m/s$

4) Calculation:

a) $C$ isotope of carbon undergoes positron emission.

The balanced equation for decay of $C$,

$\mathrm{C}\to \mathrm{B}+\mathrm{{\rm B}}$

b) The binding energy is calculated using Albert Einstein equation and exact mass of $C$ is $1.82850\times {10}^{-26} kg$

To calculate the mass of the nucleons, we have to subtract the mass of 6 electrons from exact mass

$C: (1.82850\times {10}^{-26} kg) – 6(9.10939\times {10}^{-31}kg)$

$C: = 1.827953\times {10}^{-26} kg$

Now, we calculate the mass of neutrons and protons present in the one atom of $C$

$mass of C\mathrm{ }: (6 protons \times \frac{1.67262\times {10}^{-27}kg}{protons}) + (5 neutrons \times \frac{1.67493\times {10}^{-27}kg}{neutrons})$

$mass of C : (1.003572\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g})\mathrm{ }+\mathrm{ }(8.37465\times {10}^{-27}\mathrm{ }\mathrm{k}\mathrm{g})$

$mass of C=1.841037 \times {10}^{-26} kg$

Now, calculate the mass defect $(∆\mathrm{m})$ from the exact mass of the nucleus of $C$ and its subatomic particles.

$\left(∆\mathrm{m}\right) =\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s}\mathrm{ }–\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{o}\mathrm{n}\mathrm{s}$

$\left(∆\mathrm{m}\right)=1.841037 \times {10}^{-26} kg- 1.827953\times {10}^{-26} kg$

$\left(∆\mathrm{m}\right)=1.308356 \times {10}^{-28}kg$

Now, calculate the binding energy (BE) by using Albert Einstein equation

$BE =(∆m)c^2$

$BE =1.308356 \times {10}^{-28}kg \times {\left(2.998 \times {10}^8\frac{m}{s}\right)}^2$

$BE =1.17595 \times {10}^{-11}kg \frac{m^2}{s^2} \times \frac{1 J}{kg \frac{m^2}{s^2} }=1.176 \times {10}^{-11}J$

$BE =1.176 \times {10}^{-11}J$

Conclusion:

We determined the balanced nuclear equation and calculated the binding energy from the exact mass and mass defect.