Atoms & Nuclei

JEE Main level, numerical answer, medium

4 of 239

Question 4 of 239

The mass defect in a particular reaction is 0.4g0.4g. The amount of energy liberated is n×107kWhn\times {10}^{7}kWh, where n=n= _____. (speed of light =3×108ms−1=3\times {10}^{8}m{s}^{-1})

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Answer: 1

Energy liberated is given by, E=∆mc2E=∆m{c}^{2}
=(0.4×10−3)×(3×108)2=\left(0.4\times {10}^{-3}\right)\times {\left(3\times {10}^{8}\right)}^{2}
=3600×107kWs=3600\times {10}^{7}kWs
=3600×1073600kWh=1×107kWh=\dfrac{3600\times {10}^{7}}{3600}kWh=1\times {10}^{7}kWh

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