A uranium reactor generates thermal energy at a rate of 300 MW . Determine the amount of ^ 235 U consumed every second, given that the average energy released per fission is 200 MeV.

Correct answer: D. 3.7 mg Power generated, P = 300 MW = 300 × 10^6 J/s Energy released per fission, E = 200 MeV = 200 × 10^6 × 1.6 × 10^ -19 J = 3.2 × 10^ -11 J Number of fissions per second, n = (P)/(E) = 300 × 10^6 3.2 × 10^ -11 = 9.375 × 10^ 18 Mass of one ^ 235 U atom, m = 235 6.022 × 10^ 23 g Mass of ^ 235 U consumed per second = n × m = 9.375 × 10^ 18 × 235 6.022 × 10^ 23 g Mass consumed 3.65 × 10^ -3 g = 3.65 mg 3.7 mg

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: The Nucleus

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Nucleus

A uranium reactor generates thermal energy at a rate of 300 MW ....

A uranium reactor generates thermal energy at a rate of

300 \text{ MW}

. Determine the amount of ^{235}\text{U}

consumed every second, given that the average energy released per fission is

200 \text{ MeV}.

Options

A $3.7 \text{ \mu g}$
B $3.7 \text{ g}$
C $3.7 \text{ kg}$
D $3.7 \text{ mg}$

Correct answer

D. $3.7 \text{ mg}$

Step-by-step solution

Power generated,

P = 300 \text{ MW} = 300 \times 10^6 \text{ J/s}

Energy released per fission,

E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}

Number of fissions per second,

n = \dfrac{P}{E} = \dfrac{300 \times 10^6}{3.2 \times 10^{-11}} = 9.375 \times 10^{18}

Mass of one ^{235}\text{U}

atom,

m = \dfrac{235}{6.022 \times 10^{23}} \text{ g}

<br><br>Mass of ^{235}\text{U}

consumed per second =

n \times m = 9.375 \times 10^{18} \times \dfrac{235}{6.022 \times 10^{23}} \text{ g}

\Rightarrow \text{Mass consumed} \approx 3.65 \times 10^{-3} \text{ g} = 3.65 \text{ mg} \approx 3.7 \text{ mg}

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