Question

A nuclear power plant produces $800 MW$ of electricity with an overall efficiency of $0.32$ . The fission reaction taking place in the core of the reactor is:

${}_{0}^{1}n + {}_{92}^{235}U \to {}_{54}^{140}{Xe} + {}_{38}^{94}{Sr} + 2 {}_{0}^{1}n$

(i) Using the masses provided below show that the energy released in one fission is about $180 MeV$

${}_{0}^{1}n = 1.009 u, {}_{92}^{235}U = 235.044, {}_{54}^{140}{Xe} = 139.992, {}_{38}^{94}{Sr} = 93.915$

(ii) Estimate the specific energy of $U - 235$

(iii)Show that the mass of uranium undergoing fission in one year is about $1500 kg$

Accepted Answer

The given nuclear reaction is ${}_{0}^{1}n + {}_{92}^{235}U \to {}_{54}^{140}{Xe} + {}_{38}^{94}{Sr} + 2 {}_{0}^{1}n$

Balancing the masses of the reactants and the products of the reaction, we have

$1.009 + 235.044 \to 139.922 + 93.915 + 2 \times 1.009 + Q$

We have the mass defect $△ M = 236.053 - 235.855 = 0.198 u$

As $1 u = 931 Mev$ , we get the energy in one fission reaction as

$△ M \times 931 = 0.198 \times 931 = 184 Mev$ ..........................................(1)

Now specific energy is energy per unit mass

The number of atoms in $1 kg U - 235$ is

$N = \frac{6.03 \times 10^{23}}{235} \times 1000$

The specific energy = energy contained in $1 kg$ $\frac{6.03 \times 10^{23}}{235} \times 1000 \times 931 = 4.72 \times 10^{27} MeV {kg}^{- 1}$

Or $7.55 \times 10^{13} J / kg$

(iii) The power plant produces power at the rate of $800 MW$

Efficiency= $32 %$ . Therefore, total input required = $\frac{800}{0.32} = 2500 MW = 2500 MJ / \sec$

Therefore, the energy produced per year= $2500 \times 10^{6} \times 360 \times 24 \times 60 \times 60 = 7.78 \times 10^{16} J$ .

Mass of uranium required for this energy= $\frac{Energy required}{sp . energy} = \frac{7.78 \times 10^{16} J}{7.55 \times 10^{13} J / kg} = 1050 kg$

K A Tsokos Solutions for Chapter: Energy Production, Exercise 3: Exam-style questions

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