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WAEC Physics 2015 Paper

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Question : 46

Total: 50

in a nuclear fission process, the mass defect was 1.2 x 10^-5kg. Calculate the energy released. [speed of the light in vacuum = c = 3.0 x 10⁸ms^-1]

1.08 x 10¹¹J
5.20 x 10¹¹J
1.08 x 10¹²J
2.50 x 10¹³J

Solution:

Concept:

The energy released in nuclear fission is found using Einstein's mass–energy equivalence formula

E = \Delta m c^2

, where

\Delta m

is the mass defect.

Explanation:

Write down the given values:

\Delta m = 1.2 \times 10^{-5}\,\text{kg}

and

c = 3.0 \times 10^{8}\,\text{m s}^{-1}

.

Substitute into

E = \Delta m c^2

.

E = (1.2 \times 10^{-5}) \times (3.0 \times 10^{8})^2

(3.0 \times 10^{8})^2 = 9.0 \times 10^{16}

E = 1.2 \times 10^{-5} \times 9.0 \times 10^{16}

E = 10.8 \times 10^{11}

E = 1.08 \times 10^{12}\,\text{J}

Answer:

C.

1.08 \times 10^{12}\,\text{J}

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