Determine the energy released by 1 g of natural uranium, given that each fission event releases 200 MeV and the fissionable isotope ^ 235 U constitutes 0.7\% by weight in natural uranium.

Correct answer: B. 5.7 × 10^8 J Mass of ^ 235 U in 1 g of natural uranium is m = (0.7)/(100) × 1 g = 0.007 g . Number of atoms of ^ 235 U in 0.007 g is N = (0.007)/(235) × 6.022 × 10^ 23 1.794 × 10^ 19 . Energy released per fission is E_1 = 200 MeV = 200 × 10^6 × 1.6 × 10^ -19 J = 3.2 × 10^ -11 J . Total energy released is E = N × E_1 = 1.794 × 10^ 19 × 3.2 × 10^ -11 J 5.7 × 10^8 J .

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

Determine the energy released by 1 g of natural uranium, given that...

Determine the energy released by

1 \text{ g}

of natural uranium, given that each fission event releases

200 \text{ MeV}

and the fissionable isotope ^{235}\text{U}

constitutes

0.7\% by weight in natural uranium.

Options

A $8.2 \times 10^{10} \text{ J}$
B $5.7 \times 10^8 \text{ J}$
C $2.8 \times 10^8 \text{ J}$
D $5.7 \times 10^{10} \text{ J}$

Correct answer

B. $5.7 \times 10^8 \text{ J}$

Step-by-step solution

Mass of ^{235}\text{U}

in

1 \text{ g}

of natural uranium is

m = \dfrac{0.7}{100} \times 1 \text{ g} = 0.007 \text{ g}

.<br><br>Number of atoms of ^{235}\text{U}

0.007 \text{ g}

N = \dfrac{0.007}{235} \times 6.022 \times 10^{23} \approx 1.794 \times 10^{19}

.

Energy released per fission is

E_1 = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}

.

Total energy released is

E = N \times E_1 = 1.794 \times 10^{19} \times 3.2 \times 10^{-11} \text{ J} \approx 5.7 \times 10^8 \text{ J}

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