Answer
(a) $2.4\times 10^{18} reactions$
(b) $0.93mg$
Work Step by Step
(a) We know that
$E=P_{bulb}t$
$\implies E=(120W)(2.5d)(86,400s/d)=26MJ$
Now $N_{reaction}=\frac{E}{\epsilon E_{fission}}$
We plug in the known values to obtain:
$N=\frac{25.9MJ}{(0.32)(212MeV)}$
$N=2.4\times 10^{18} reactions$
(b) We can find the required mass as
$m=N_{reactions}M_U$
$\implies m=(2.39\times 10^{18})(235.043925u)(1.66\times 10^{-27}Kg/u)$
$m=9.3\times 10^{-7}kg=0.93mg$
