Energy released during the fission of one Uranium-235 nucleus is 200MeV. Energy released by the fission of 500gm of U-235 nuclei will be about

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1.6 × 10³¹ MeV

3.5 × 10²⁰ MeV

2.8 × 10²⁶ MeV

6.6 × 10²⁴ MeV

answer is C.

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Detailed Solution

$\frac{n}{N_{A}} = \frac{m}{M} \Rightarrow n = \frac{{mN}_{A}}{M}$
No. of atoms $n = \frac{500 \times 6 \times 10^{23}}{235} = 12 .76 \times 10^{23}$
E¹ = n.E
E¹ = 12.76 ×10²³ × 200 = 2.552 × 10²⁶MeV

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