Chapter 21, Problem 21.50QA

Interpretation Introduction

To explain:

Which isotope has larger binding energy:$B and B$?

Answer to Problem 21.50QA

Solution:

Isotope $B$ has larger binding energy than $B.$

Explanation of Solution

1) Concept:

To calculate the binding energy of the nuclei of two isotopes, we need to calculate the mass defect of the nuclei. From the exact mass of the isotopes, we will calculate the exact mass of each isotope. From the mass of each isotope, we will calculate the mass of the nucleons. Calculate the mass of the nucleus by using the mass of photons and the mass of the neutrons. Calculate the mass defect from the mass of nucleus and the mass of nucleons. We will calculate the binding energy from the mass defect.

2) Formula:

Albert Einstein equation for the binding energy is

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

3) Given:

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

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

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

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

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

vi) The exact mass of $B.$ is $10.0129amu$.

vii) The exact mass of $B$. is $11.0093 amu$.

Calculation:

First, we convert the exact mass of each isotope from $amu to kg.$

$B: 10.0129 amu \times \frac{1.66054\times {10}^{-27}kg}{amu} = 1.662682\times {10}^{-26} kg$

$B: 11.0093 amu \times \frac{1.66054\times {10}^{-27}kg}{amu} = 1.828138\times {10}^{-26} kg$

To calculate the mass of the nucleons, we have to subtract the mass of $5$ electrons from the exact mass.

$B: \left(1.662682\times {10}^{-26} kg\right)- 5\left(9.10939\times {10}^{-31} kg\right)=1.662227\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g}$

$B: \left(1.828138\times {10}^{-26}kg\right)- 5\left(9.10939\times {10}^{-31} kg\right)=1.827676\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g}$

Now, we calculate the mass of the neutrons and the protons present in one atom of each isotope.

$B: \left(5 protons \times \frac{1.67262\times {10}^{-27}kg}{protons}\right)+ \left(5 neutrons \times \frac{1.67493\times {10}^{-27}kg}{neutrons}\right)=\mathrm{ }1.673775\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g}$

$Cl: \left(5 protons \times \frac{1.67262\times {10}^{-27}kg}{protons}\right)+ \left(6 neutrons \times \frac{1.67493\times {10}^{-27}kg}{neutrons}\right)=\mathrm{ }1.841268\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g}$

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

$(∆m) = mass of nucleus – mass of nucleons$

$B: ∆m= 1.673775\times {10}^{-26} kg – 1.662227\times {10}^{-26} kg=1.1548 \times {10}^{-28} kg$

$B: ∆m= 1.841268\times {10}^{-26} kg – 1.827676\times {10}^{-26} kg=1.3585 \times {10}^{-28} kg$

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

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

$B: BE =1.1548 \times {10}^{-28} kg \times {\left(2.998 \times {10}^8\frac{m}{s}\right)}^2$

$B: BE = 1.03793 \times {10}^{-11} kg \frac{m^2}{s^2} \times \frac{1 J}{kg \frac{m^2}{s^2} }=1.0379 \times {10}^{-11} J$

$B: BE =1.3585 \times {10}^{-28} kg\mathrm{ } \times {\left(2.998 \times {10}^8\frac{m}{s}\right)}^2$

$B: BE = 1.22102 \times {10}^{-11} kg \frac{m^2}{s^2} \times \frac{1 J}{kg \frac{m^2}{s^2} }=1.2210 \times {10}^{-11} J$

Hence, the binding energy of isotope $B$ is larger than that of isotope $B$.

Conclusion:

We calculated the binding energy of each isotope from the exact mass and the mass defect.