Answer
(a) The mass of a single atom of $U^{235}$
$m_0$= $(235u)(1.661 \times 10^{-27} kg/u)$ = $3.90 \times 10^{-25} kg$
So the number of atoms in $m=1.0 kg$ is
$N$ =$m/m_0$ = $(1.0 kg)/(3.90 \times10^{-25} kg)$ = $2.56 \times10^{24}$
An alternate approach (but essentially the same once) would be using Eq. 42-21.
(b):
The energy released by N fission events is given by $E=NQ$, where $Q$ is the energy released in each event. For $1.0 kg$ of $U^{235}$ we have:
$E$=$(2.56 \times10^{24}) (200 \times 10^6 eV)(1.6 \times 10^{-19} J/eV)$= $8.19 \times 10^{13} J$
(c):
If P is the power requirement of the lamp, then:
$t$=$E/P$= $8.19 \times 10^{13} J (100 W)$ = $8.19 \times 10^{11} s$
Also we know: $y/s$ = $3.156 \times 10^7$
So the answer is :
$(8.19 \times 10^{11} s) \times (3.156 \times 10^7)$ =$2.6 \times 10^{4} y$
Work Step by Step
(a) The mass of a single atom of $U^{235}$
$m_0$= $(235u)(1.661 \times 10^{-27} kg/u)$ = $3.90 \times 10^{-25} kg$
So the number of atoms in $m=1.0 kg$ is
$N$ =$m/m_0$ = $(1.0 kg)/(3.90 \times10^{-25} kg)$ = $2.56 \times10^{24}$
An alternate approach (but essentially the same once) would be using Eq. 42-21.
(b):
The energy released by N fission events is given by $E=NQ$, where $Q$ is the energy released in each event. For $1.0 kg$ of $U^{235}$ we have:
$E$=$(2.56 \times10^{24}) (200 \times 10^6 eV)(1.6 \times 10^{-19} J/eV)$= $8.19 \times 10^{13} J$
(c):
If P is the power requirement of the lamp, then:
$t$=$E/P$= $8.19 \times 10^{13} J (100 W)$ = $8.19 \times 10^{11} s$
Also we know: $y/s$ = $3.156 \times 10^7$
So the answer is :
$(8.19 \times 10^{11} s) \times (3.156 \times 10^7)$ =$2.6 \times 10^{4} y$
