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Q.How many energy will be released in joule due to mass defect of 1 gram?

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2018Subjective· 1mImportance★★★★★
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E=mc2=9×1013 JE = mc^2 = 9\times10^{13}\ \text{J} for a 1 gram mass defect.

Concept. Einstein's mass–energy equivalence states that a mass defect Δm\Delta m is released as energy E=Δm c2E=\Delta m\,c^2.

Steps.

Δm=1 g=1×10−3 kg,c=3×108 m s−1\Delta m = 1\,\text{g} = 1\times10^{-3}\,\text{kg}, \qquad c = 3\times10^{8}\,\text{m s}^{-1}

E=Δm c2=(10−3)×(3×108)2=10−3×9×1016E = \Delta m\, c^2 = (10^{-3})\times(3\times10^{8})^2 = 10^{-3}\times 9\times10^{16} …

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