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Question

Physics Question on Nuclear physics

The energy in MeVMeV is released due to transformation of 1kg1\, kg mass completely into energy, is (c=3×108m/s)\left(c=3 \times 10^{8} m / s \right)

A

7.625×10MeV7.625\times 10\,MeV

B

10.5×1029MeV 10.5\times 10^{29}\,MeV

C

2.8×1028MeV2.8\times 10^{-28}\,MeV

D

5.625×1029MeV5.625\times 10^{29}\,MeV

Answer

5.625×1029MeV5.625\times 10^{29}\,MeV

Explanation

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} J
=9×10161.6×1013MeV=\frac{9 \times 10^{16}}{1.6 \times 10^{-13}} MeV
( since 1MeV=1.6×1013J)\left(\text { since } 1\, MeV =1.6 \times 10^{-13} J \right)
=5.625×1029MeV=5.625 \times 10^{29}\, MeV