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Angle Sum Property

The binding e...

Question

The binding energy of deutron $_{1} H^{2}$ is $1.112 M e V$ per nucleon and an alpha particle $_{2} H e^{4}$ has a binding energy of $7.047 M e V$ per nucleon. Then, in the fusion rection
$_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + Q$ , the energy $Q$ released is
(take mass of $_{1} H^{2} = 2.01478 u$ ;
mass of $_{2} H e^{4} = 4.00338 u$ )

A

1.8 M e V

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B

11.9 M e V

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C

23.8 M e V

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D

931 M e V

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Solution

The correct option is C $23.8 M e V$
Mass of $_{1} H^{2} = 2.01478 u$
Mass of 2 deuterium $= 2 \times 2.01478 = 4.02956 u$
Mass of $_{2} H e^{4} = 4.00338 u$

Energy equivalent to 2 $_{1} H^{2} = 4.02956 \times 1.112 M e V = 4.48 M e V$
Energy equivalent to 2 $_{2} H e^{4} = 4.00388 \times 7.047 M e V = 28.21 M e V$

Energy released $= 28.21 - 4.48 = 23.73 M e V \approx 23.8 M e V$

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Similar questions

Q. The binding energy of deutron

_{1} H^{2}

1.112 M e V

per nucleon and an alpha particle

_{2} H e^{4}

has a binding energy of

7.047 M e V

per nucleon. Then, in the fusion rection

_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + Q

Q

released is
(take mass of

_{1} H^{2} = 2.01478 u

_{2} H e^{4} = 4.00338 u

Q. The amount of energy released in the fusion of

_{1} H^{2}

_{2} H e^{4}

nucleus will be :
[Binding energy per nucleon of

_{1} H^{2} = 1.1 M e V

_{2} H e^{4} = 7 M e V

]

Q. Binding energy per nucleon of

_{1} H^{2}

_{2} H e^{4}

1.1 M e V

7.0 M e V

, respectively. Energy released in the process

_{1} H^{2} +_{1} H^{2} ⟶_{2} H e^{4}

Q. The binding energies per nucleon for deutron

(_{1} H^{2})

(_{2} H e^{4})

1.1

7.0

MeV respectively. Calculate the energy released, when two deutrons fuse to form a helium nucleus

(_{2} H e^{4})

Q. The binding energy per nucleon for a

deuteron

α - particle

x_{1}

x_{2}

respectively. The energy

Q

released in the reaction

_{1} H^{2} +_{1} H^{2} \to_{2} {He}^{4} + Q,

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