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JIPMER 2004
Exam
JIPMER
Year
2004
Subject
Physics
Chapter
Nuclear Physics

JIPMER 2004 · Physics · Nuclear Physics

The energy in MeV is released due to transformation of 1 kg mass...

The energy in MeV is released due to transformation of 1 kg mass completely into energy, is (c=3×108 m/s)\left(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)

Options

  1. A7.625×10MeV7.625 \times 10 \mathrm{MeV}
  2. B10.5×1029MeV10.5 \times 10^{29} \mathrm{MeV}
  3. C2.8×10−28MeV2.8 \times 10^{-28} \mathrm{MeV}
  4. D5.625×1029MeV5.625 \times 10^{29} \mathrm{MeV}

Correct answer

D. 5.625×1029MeV5.625 \times 10^{29} \mathrm{MeV}

Step-by-step solution

According to Einstein's mass energy equivalence E=mc2=1×(3×108)2=9×1016JE=m c^2=1 \times\left(3 \times 10^8\right)^2=9 \times 10^{16} \mathrm{ J}=9×10161.6×10−13MeV (since 1MeV=1.6×10−13J ) \begin{aligned} & =\frac{9 \times 10^{16}}{1.6 \times 10^{-13}} \mathrm{MeV} \\ & \quad \text{ (since } 1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{ J} \text{ ) }\end{aligned}=5.625×1029MeV=5.625 \times 10^{29} \mathrm{MeV}
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