The energy in MeV released due to transformation of 1 kg mass completely into energy, is (c = 3 × 10⁸m/s)

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7.625×10⁹MeV

2.8×10^–28MeV

5.625×10²⁹MeV

29×10⁵MeV

answer is D.

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Detailed Solution

$\begin{array}{l} E = {mc}^{2} = 9 \times 10^{16} J \\ = \frac{9 \times 10^{16}}{1 .6 \times 10^{- 19}} eV = 5 .625 \times 10^{35} eV \\ E = 5 .625 \times 10^{29} MeV \end{array}$

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