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:

2 m + \frac{E}{c^{2}}

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\frac{E}{{m c}^{2}}

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m + \frac{E}{c^{2}}

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2 m - \frac{E}{c^{2}}

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Solution

The correct option is D $2 m - \frac{E}{c^{2}}$
Mass lost must be converted to energy.
By Einstein's mass energy equivalence:

$E = Δ m c^{2}$

$E = (2 m - M) c^{2}$ ; where $M$ is mass of Helium nucleus.

Hence, $M = 2 m - \frac{E}{c^{2}}$

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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:

The correct option is D 2m−Ec2Mass lost must be converted to energy. By Einstein's mass energy equivalence:E=Δmc2E=(2m−M)c2; where M is mass of Helium nucleus.Hence, M=2m−Ec2

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:

The correct option is D $2 m - \frac{E}{c^{2}}$
Mass lost must be converted to energy.
By Einstein's mass energy equivalence:

$E = Δ m c^{2}$

$E = (2 m - M) c^{2}$ ; where $M$ is mass of Helium nucleus.

Hence, $M = 2 m - \frac{E}{c^{2}}$