Latihan Soal RADIOISOTOP
Suatu reaksi fisi menghasilkan energi 210 MeV tiap fisi. Berapa banyak fisi yang terjadi tiap sekon bila reaktor menghasilkan daya 130 MW
Diketahui :
\begin{align} E_{f} &= 210 \ MeV = 210 \times 10^{6} \ eV = 2,1 \times 10^{8} eV \nonumber\\ &= 2,1 \times10^{8} \times 1,6 \times 10^{-19} J =3,36\times10^{-11} J\nonumber \\ t &= 1 \ sekon \nonumber\\ P &= 130 \ MW = 130 \times10^{6} \ W = 1,3 \times10^{8} \ W \nonumber \end{align}
Ditanya :
\begin{align} N = ...? \nonumber \end{align}
Jawab :
\begin{align} E &= P . t \nonumber\\ &= 1,3 \times10^{8} \ W . 1 \ sekon \nonumber\\&=1,3 \times10^{8} \ J \nonumber\\N &= \frac{E}{E_{f}} \nonumber\\ &= \frac{1,3 \times10^{8} \ J}{3,36\times10^{-11}} \nonumber\\ &= 3,87 \times 10^{18} \ fisi \nonumber \end{align}
Diketahui :
\begin{align} m_{batu \ bara} &= 1 \ kg \nonumber\\ E_{(1 \ kg)} &= 3\times 10^{7} \ J \nonumber\\ E_{f \ uranium} &= 200 \ MeV = 2 \times 10^{8} eV \nonumber\\ &= 2 \times 10^{8} \times 1,6 \times 10^{-19} J =3,2\times10^{-11} J\nonumber \\ Ar &= 235 \nonumber \end{align}
Ditanya :
\begin{align} m_{batu \ bara} = ...? \nonumber \end{align}
Jawab :
\begin{align} E_{total batu bara} &= E_{uranium}\nonumber\\ m . E_{1 \ kg} &= N. E_{f \ 1 \ kg} \nonumber\\ m . E_{1 \ kg} &= \frac{m. N_{0}}{Ar} . E_{f \ 1 \ kg} \nonumber\\ m. 3\times 10^{7} &= \frac{1.000 \times 6,02\times10^{23}}{235}. 3,2\times10^{-11} \nonumber\\ m. 3\times 10^{7} &= 8,197 \times 10^{13} \nonumber\\ m &= \frac{8,197 \times 10^{13} }{3\times10^{7}} \nonumber\\ m &= 2,73 \times 10^{6} \ kg \nonumber \end{align}
Air yang masuk reaktor mempunyai suhu 216^o C dan meninggalkan reaktor dengan suhu 287^o C (tekanan reaktor sangat tinggi sehingga tidak terjadi penguapan). Reaktor melepas panas sebanyak 5,6x10^{9} J. Asumsikan kapasitas panas air tersebut 4.420 J/kg^oC, hitung massa air yang harus disediakan untuk mendinginkan reaktor.
Diketahui :
\begin{align} T_{1} &=216^{o}C \nonumber\\ T_{2}&= 287^{o} C \nonumber\\ \Delta T &= 287 - 216 = 71^{o}C \nonumber\\ Q_{lepas} &=5,6x10^{9} \ J \nonumber\\ c_{air} &= 4.420 J/kg^{o}C \nonumber \end{align}
Ditanya :
\begin{align} m_{air} = ...? \nonumber \end{align}Jawab :
\begin{align} m_{air} \times c_{air} \times (\Delta T) &= Q_{lepas} \nonumber\\ m_{air}&= \frac{Q_{lepas}}{c_{air} \times (\Delta T)} \nonumber\\ &=\frac{5,6x10^{9} }{4.420 \times 71} \nonumber\\ &= 1.7845 \ kg \nonumber \end{align}
Diketahui :
\begin{align}\eta &=25 \% =\frac{1}{4} \nonumber\\ P_{output}&= 800 \ MW = 8 \times10^{8} W \nonumber\\ E_{f} &= 200 \ MeV = 2 \times10^{8} \ eV \nonumber\\ &=2 \times 10^{8} \times 1,6 \times10^{-19} = 3,2 \times10^{-11} J \nonumber\\ t &= 1 \ tahun = 365 \times 24 \times 60 \times 60 = 31.536.000 \ detik\nonumber \end{align}
Jawab :
Jumlah inti unsur uranium :
\begin{align} N \times E_{f} &= E_{input} \nonumber\\ N \times E_{f} &= P_{input} \times t \nonumber\\ N \times 3,2 \times10^{-11} &= 32 \times10^{8} \times 31.536.000 \nonumber\\ N &= 3,1536 \times 10^{27}\nonumber \end{align}
massa uranium :
\begin{align} N &= \frac {m \times N_{0}}{Ar} \nonumber\\ m &= \frac {N \times Ar}{ N_{0}} \nonumber\\ &= \frac{3,1536 \times 10^{27} \times 235}{6,02 \times10^{23}}\nonumber\\ &= 1.231.056 \ kg \nonumber \end{align}
Diketahui :
\begin{align} P = ...?\nonumber \end{align}
Jawab :
\begin{align}P &= \frac{E}{t} \nonumber\\ &= \frac{N \times E_{f}}{t} \nonumber\\ &= \frac{3 \times 10^{19} \times 3,2 \times 10^{-11}}{1} \nonumber\\ &=960.000.000 \ W \nonumber\\ &=960 \ MW \nonumber \end{align}
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