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Two deuterium nuclei each of mass 'm' fuse together, to form a Helium nucleus, releasing an energy E . If 'c' is the velocity of light , the mass of Helium nucleus formed will be :

$\begin {array} {1 1} (1)\;2m+ \large\frac{E}{e^2} & \quad (2)\;\large\frac{E}{me^2} \\ (3)\;m+\large\frac{E}{e^2} & \quad (4)\;2m-\large\frac{E}{e^2} \end {array}$

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(4) $ 2m - \large\frac{E}{e^2}$
answered Nov 7, 2013 by pady_1

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