Answer :
We begin by noting the following data:
- The observed atomic mass of a [tex]${}^{235}U$[/tex] atom is
[tex]$$235.043923 \text{ u}.$$[/tex]
- The conversion factor from atomic mass units to kilograms is
[tex]$$1\, \text{u} = 1.66053906660 \times 10^{-27}\, \text{kg}.$$[/tex]
- One mole of any substance contains Avogadro's number of atoms,
[tex]$$N_A = 6.02214076 \times 10^{23}\, \text{atoms/mol}.$$[/tex]
- The speed of light is
[tex]$$c = 299792458\, \text{m/s}.$$[/tex]
- The energy needed to power a 100-watt lightbulb for 1 hour is
[tex]$$360\, \text{kJ} = 360 \times 10^3\, \text{J}.$$[/tex]
- One year is taken to be
[tex]$$365.242\, \text{days},$$[/tex]
so the number of hours in one year is
[tex]$$365.242 \times 24 \approx 8765.808\, \text{hours}.$$[/tex]
We want to determine for how many years the energy available from the complete conversion of one mole of [tex]${}^{235}U$[/tex] into energy (using Einstein’s equation [tex]$E = mc^2$[/tex]) can power the lightbulb.
Step 1. Mass of a [tex]${}^{235}U$[/tex] Atom
Convert the atomic mass into kilograms:
[tex]$$
m_{\text{atom}} = 235.043923 \times 1.66053906660 \times 10^{-27} \, \text{kg} \approx 3.9029961650842226 \times 10^{-25}\, \text{kg}.
$$[/tex]
Step 2. Mass of One Mole of [tex]${}^{235}U$[/tex]
Multiply the mass of one atom by Avogadro's number:
[tex]$$
m_{\text{mole}} = m_{\text{atom}} \times N_A \approx 3.9029961650842226 \times 10^{-25}\, \text{kg} \times 6.02214076 \times 10^{23} \approx 0.23504392291877385\, \text{kg}.
$$[/tex]
Step 3. Total Energy Available from the Mass Defect
Using Einstein’s mass–energy equivalence:
[tex]$$
E = m_{\text{mole}} c^2.
$$[/tex]
Substitute the values:
[tex]$$
E \approx 0.23504392291877385\, \text{kg} \times (299792458\, \text{m/s})^2 \approx 2.1124694295386536 \times 10^{16}\, \text{J}.
$$[/tex]
Step 4. Hours the Lightbulb Can Run
The lightbulb requires [tex]$360 \times 10^3\, \text{J}$[/tex] per hour. Thus, the number of hours the bulb can be powered by the available energy is:
[tex]$$
\text{Total hours} = \frac{E}{360 \times 10^3 \, \text{J}} \approx \frac{2.1124694295386536 \times 10^{16}\, \text{J}}{360 \times 10^3\, \text{J}} \approx 5.867970637607371 \times 10^{10}\, \text{hours}.
$$[/tex]
Step 5. Convert Hours to Years
Dividing by the number of hours in a year gives:
[tex]$$
\text{Total years} = \frac{5.867970637607371 \times 10^{10}\, \text{hours}}{8765.808\, \text{hours/year}} \approx 6.694158299619807 \times 10^{6}\, \text{years}.
$$[/tex]
Thus, the energy corresponding to the mass defect of one mole of [tex]${}^{235}U$[/tex] could keep a 100-watt incandescent lightbulb burning for approximately
[tex]$$
\boxed{6.69 \times 10^6 \text{ years}}.
$$[/tex]
- The observed atomic mass of a [tex]${}^{235}U$[/tex] atom is
[tex]$$235.043923 \text{ u}.$$[/tex]
- The conversion factor from atomic mass units to kilograms is
[tex]$$1\, \text{u} = 1.66053906660 \times 10^{-27}\, \text{kg}.$$[/tex]
- One mole of any substance contains Avogadro's number of atoms,
[tex]$$N_A = 6.02214076 \times 10^{23}\, \text{atoms/mol}.$$[/tex]
- The speed of light is
[tex]$$c = 299792458\, \text{m/s}.$$[/tex]
- The energy needed to power a 100-watt lightbulb for 1 hour is
[tex]$$360\, \text{kJ} = 360 \times 10^3\, \text{J}.$$[/tex]
- One year is taken to be
[tex]$$365.242\, \text{days},$$[/tex]
so the number of hours in one year is
[tex]$$365.242 \times 24 \approx 8765.808\, \text{hours}.$$[/tex]
We want to determine for how many years the energy available from the complete conversion of one mole of [tex]${}^{235}U$[/tex] into energy (using Einstein’s equation [tex]$E = mc^2$[/tex]) can power the lightbulb.
Step 1. Mass of a [tex]${}^{235}U$[/tex] Atom
Convert the atomic mass into kilograms:
[tex]$$
m_{\text{atom}} = 235.043923 \times 1.66053906660 \times 10^{-27} \, \text{kg} \approx 3.9029961650842226 \times 10^{-25}\, \text{kg}.
$$[/tex]
Step 2. Mass of One Mole of [tex]${}^{235}U$[/tex]
Multiply the mass of one atom by Avogadro's number:
[tex]$$
m_{\text{mole}} = m_{\text{atom}} \times N_A \approx 3.9029961650842226 \times 10^{-25}\, \text{kg} \times 6.02214076 \times 10^{23} \approx 0.23504392291877385\, \text{kg}.
$$[/tex]
Step 3. Total Energy Available from the Mass Defect
Using Einstein’s mass–energy equivalence:
[tex]$$
E = m_{\text{mole}} c^2.
$$[/tex]
Substitute the values:
[tex]$$
E \approx 0.23504392291877385\, \text{kg} \times (299792458\, \text{m/s})^2 \approx 2.1124694295386536 \times 10^{16}\, \text{J}.
$$[/tex]
Step 4. Hours the Lightbulb Can Run
The lightbulb requires [tex]$360 \times 10^3\, \text{J}$[/tex] per hour. Thus, the number of hours the bulb can be powered by the available energy is:
[tex]$$
\text{Total hours} = \frac{E}{360 \times 10^3 \, \text{J}} \approx \frac{2.1124694295386536 \times 10^{16}\, \text{J}}{360 \times 10^3\, \text{J}} \approx 5.867970637607371 \times 10^{10}\, \text{hours}.
$$[/tex]
Step 5. Convert Hours to Years
Dividing by the number of hours in a year gives:
[tex]$$
\text{Total years} = \frac{5.867970637607371 \times 10^{10}\, \text{hours}}{8765.808\, \text{hours/year}} \approx 6.694158299619807 \times 10^{6}\, \text{years}.
$$[/tex]
Thus, the energy corresponding to the mass defect of one mole of [tex]${}^{235}U$[/tex] could keep a 100-watt incandescent lightbulb burning for approximately
[tex]$$
\boxed{6.69 \times 10^6 \text{ years}}.
$$[/tex]
