Chapter 14, Problem 14.43PAE

Interpretation Introduction

Interpretation: To determine the how many number of years keep burning the bulb by mass defect energy of 1mole ${}^{\text{14}}\text{C}$ keep a 100-W light bulb burning.

Concept introduction: Mass defect:

The nucleus of each atom (aside from ${}_{\text{1}}^{\text{1}}\text{H}$ ) has a mass lower than expected from adding the masses of its neutrons and protons.

Mass and energy are related by following equation.

$\mathrm{E}=m{c}^2$

E= Energy

m= mass

c = speed of light.

Loss of mass and liberation of energy are related by Einstein’s equations.

$\begin{array}{l}{\text{E = (Δm)c}}^{\text{2}}\\ \text{Δm = Mass defect}\\ \text{c = Velocity of light}\end{array}$

Answer to Problem 14.43PAE

Solution:

The mass defect energy keeps the burning up to 3221years.

Explanation of Solution

Formula used:

$\begin{array}{l}{\text{E = (Δm)c}}^{\text{2}}\\ \text{Δm = Mass defect}\\ \text{c = Velocity of light}\end{array}$

Calculation: Let’s calculate the calculated mass of ${}^{\text{14}}\text{C}$

${}^{\text{14}}\text{C}$ have six protons and eight neutrons.

$\begin{array}{l}{}^{\text{14}}{\text{C}}_{\text{cal}}{\text{= m}}_{\text{H}}{\text{+ m}}_{\text{N}}\\ \text{Mass of proton =1}\text{.007825 u}\\ \text{Mass of neutron =1}\text{.008665 u}\end{array}$

$\begin{array}{l}\text{=6(1}\text{.007825 u) + 8(1}\text{.008665 u)}\\ \text{=6}\text{.04695 u + 8}\text{.06932 u}\\ \text{=14}\text{.11627 u}\end{array}$

The calculated mass of ${}^{\text{14}}\text{C}$ is 14.11627 u.

Let’s calculate the mass defect

$\begin{array}{l}{\text{Δm =}}^{\text{ 14}}{\text{C}}_{\text{cal}}{\text{-}}^{\text{14}}{\text{C}}_{\text{obs}}\\ {\text{Calculated mass of }}^{\text{14}}\text{C =14}\text{.11627 u}\\ {\text{Actual mass of }}^{\text{14}}\text{C =14}\text{.003242 u}\end{array}$

$\begin{array}{l}\text{=14}\text{.11627 u -14}\text{.003242 u}\\ \text{= 0}\text{.113028 u}\end{array}$

Mass defect of ${}^{\text{14}}\text{C}$ is 0.113028 u.

Let’s find the binding energy

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

$\begin{array}{c}{\text{E = (Δm)c}}^{\text{2}}\\ \text{Δm = mass defect = 0}\text{.113028 u}\\ \text{= 0}\text{.113028 u×}\left(\frac{\text{1}{\text{.66054×10}}^{\text{-27}}\text{ kg}}{\text{u}}\right)\text{ × }{\left(3\times {10}^8\right)}^{\text{2}}\\ \text{=1}{\text{.68685 × 10}}^{\text{-11}}{\text{ kg m}}^{\text{2}}{\text{s}}^{\text{-2}}\\ \text{=1}{\text{.68685 × 10}}^{\text{-11}}\text{ J}\end{array}$

Let’s calculate the energy release when ${}^{\text{14}}\text{C}$ nucleus forms from its constituent nucleon. For 1 mole of ${}^{\text{14}}\text{C}$ nuclei.

$\begin{array}{c}\text{E =}\left(\frac{\text{1}{\text{.68685×10}}^{\text{-11}}\text{ J}}{\text{1 atom}}\right)\left(\frac{\text{6}{\text{.02214×10}}^{\text{23}}\text{atoms}}{\text{1 mol}}\right)\\ \text{= 10}{\text{.15845×10}}^{\text{12}}\text{ J/mol}\\ \text{= 10}{\text{.15845×10}}^9\text{ kJ/mol}\end{array}$

Energy released by 1 mole of ${}^{\text{14}}\text{C}$ nuclei is $\text{10}{\text{.15845 × 10}}^{\text{9}}\text{ kJ/mol}$.

If it takes 360kJ of energy to keeps a 100-watt light bulb burning for 1 hour.

Assuming light bulb won’t burn out.

Let’s calculate the how many years this energy of carbon -14 could keep bulb burning.

From the given, 360 kJ will burn for 1 hour.

$\text{(10}{\text{.15845 × 10}}^{\text{9}}\text{ kJ) }\left(\frac{\text{1 hr}}{\text{360 kJ}}\right)\text{ = 2}{\text{.821791× 10}}^{\text{7}}\text{ hrs}$

Let’s converts hours into days.

$\begin{array}{c}\text{1 day = 24 hrs}\\ \text{1 year = 365 days}\end{array}$

$\begin{array}{c}\text{=(2}{\text{.821791×10}}^{\text{7}}\text{ hrs)}\left(\frac{\text{1 day}}{\text{24 hrs}}\right)\left(\frac{\text{1 year}}{\text{365 days}}\right)\\ \text{=3221 years}\end{array}$

Therefore, the mass defect energy keeps the burning up to 3221 years.

Conclusion

Therefore, the mass defect energy keeps the burning up to 3221 years.