The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV , respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

Correct answer: A. 23.6 MeV Mass of _1 H^2=2.01478 amu Mass of _2 He^4=4.00388 amu Mass of two deuterium =2 × 2.01478=4.02956 amu Energy equivalent to _2 H^2 aligned & =4.02956 × 1.2 & =4.4 MeV aligned Energy equivalent to _2 He^4=4.00388 × 7=28 MeVEnergy released =(28-4.4) =23.6 MeV

NEET 2010

Exam: NEET
Year: 2010
Subject: Physics
Chapter: Nuclear Physics

NEET 2010 · Physics · Nuclear Physics

The binding energy per nucleon in deuterium and helium nuclei are...

The binding energy per nucleon in deuterium and helium nuclei are

1.1 \mathrm{MeV}

and

7.0 \mathrm{MeV}

, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

Options

A $23.6 \mathrm{MeV}$
B $2.2 \mathrm{MeV}$
C $28.0 \mathrm{MeV}$
D $30.2 \mathrm{MeV}$

Correct answer

A. $23.6 \mathrm{MeV}$

Step-by-step solution

Mass of

{ }_1 \mathrm{H}^2=2.01478 \mathrm{amu}

Mass of

{ }_2 \mathrm{He}^4=4.00388 \mathrm{amu}

Mass of two deuterium

=2 \times 2.01478=4.02956 \mathrm{amu}

Energy equivalent to

{ }_2 \mathrm{H}^2

\begin{aligned} & =4.02956 \times 1.2 \\ & =4.4 \mathrm{MeV} \end{aligned}

Energy equivalent to

{ }_2 \mathrm{He}^4=4.00388 \times 7=28 \mathrm{MeV}\text{Energy released }=(28-4.4) =23.6 \mathrm{MeV}$

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