Chapter 21, Problem 21.48QA

Interpretation Introduction

To find:

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

b) The binding energy of $F$..

Answer to Problem 21.48QA

Solution:

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

$\mathrm{F}\to \mathrm{O}+\mathrm{{\rm B}}$

b) The binding energy of $F$ is $2.199 \times {10}^{-11}J$.

Explanation of Solution

1) Concept:

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

2) Formula:

Albert Einstein equation for the binding energy is

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

3) Given:

i) The exact mass of $F$ is $2.98915\times {10}^{-26} kg$.

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) $F$ isotope of fluorine undergoes positron emission.

The balanced equation for decay of $F$,

$\mathrm{F}\to \mathrm{O}+\mathrm{{\rm B}}$

b) Binding energy is calculated using Albert Einstein equation and exact mass of $F$ is $2.98915\times {10}^{-26} kg$

To calculate mass of the nucleons, we have to subtract the mass of 9 electrons from the exact mass.

$F: (2.98915\times {10}^{-26} kg) – 9(9.10939\times {10}^{-31}kg)$

$F: = 2.98833\times {10}^{-26} kg$

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

$mass of F\mathrm{ }: (9 protons \times \frac{1.67262\times {10}^{-27}kg}{protons}) + (9 neutrons \times \frac{1.67493\times {10}^{-27}kg}{neutrons})$

$mass of F : (1.505358\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g})\mathrm{ }+\mathrm{ }(1.507437\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g})$

$mass of F=3.012795 \times {10}^{-26} kg$

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

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

$\left(∆\mathrm{m}\right)=3.012795 \times {10}^{-26} kg- 2.98833\times {10}^{-26} kg$

$\left(∆\mathrm{m}\right)=2.446485 \times {10}^{-28}kg$

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

$BE =(∆m)c^2$

$BE =2.446485 \times {10}^{-28}kg \times {\left(2.998 \times {10}^8\frac{m}{s}\right)}^2$

$BE =2.198901 \times {10}^{-11}kg \frac{m^2}{s^2} \times \frac{1 J}{kg \frac{m^2}{s^2} }=2.199 \times {10}^{-11}J$

$BE =2.199 \times {10}^{-11}J$

Conclusion:

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