The binding energies per nucleon for a deuteron and an -particle are x 1 and x 2 respectively-What will be the energy Q released in the reaction 1 H 2 - 1 H 2 - 2 He 4 -Q

The correct option is B 4( x _ 2 x _ 1 ) The binding energy on reactant side(i.e., deutrons) is 4× x_1=4x_1 The binding energy on product side(i.e., ...

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Standard X

Physics

Nuclear Fusion

The binding e...

Question

The binding energies per nucleon for a deuteron and an $α$ -particle are $x_{1}$ and $x_{2}$ respectively.What will be the energy Q released in the reaction ${_{1} H}^{2} +_{1} H^{2} + \to_{2} {H e}^{4} + Q$

4 (x_{1} + x_{2})

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4 (x_{2} - x_{1})

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2 (x_{1} + x_{2})

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2 (x_{2} - x_{1})

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Solution

The correct option is B $4 (x_{2} - x_{1})$
The binding energy on reactant side(i.e., deutrons) is $4 \times x_{1} = 4 x_{1}$
The binding energy on product side(i.e., $α$ -particle) is $4 \times x_{2} = 4 x_{2}$
So, energy $Q$ released is $4 x_{2} - 4 x_{1} = 4 (x_{2} - x_{1})$

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Similar questions

Q. The binding energy per nucleon for a

deuteron

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x_{1}

and

x_{2}

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

Q. The binding energy per nucleon for a deuteron and an alpha particle, are

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The bond energy per nucleon for deuteron and an $α - p a r t i c l e$ are $X_{1}$ and $X_{2}$ respectively. The energy released in the reaction will be:

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

The binding energy of a deuteron $H^{2}$ is 1.15 MeV per nucleon & that of an alpha particle is 7.1 Mev per nucleon. Then in the reaction $H^{2} + H^{2} \to H e^{4} + Q,$ the energy Q released is approximately

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The binding energies per nucleon for a deuteron and an α-particle are x1 and x2 respectively.What will be the energy Q released in the reaction 1H2+1H2+→2He4+Q

The correct option is B 4(x2−x1)The binding energy on reactant side(i.e., deutrons) is 4×x1=4x1The binding energy on product side(i.e., α-particle) is 4×x2=4x2So, energy Q released is 4x2−4x1=4(x2−x1)

The binding energies per nucleon for a deuteron and an $α$ -particle are $x_{1}$ and $x_{2}$ respectively.What will be the energy Q released in the reaction ${_{1} H}^{2} +_{1} H^{2} + \to_{2} {H e}^{4} + Q$

The correct option is B $4 (x_{2} - x_{1})$
The binding energy on reactant side(i.e., deutrons) is $4 \times x_{1} = 4 x_{1}$
The binding energy on product side(i.e., $α$ -particle) is $4 \times x_{2} = 4 x_{2}$
So, energy $Q$ released is $4 x_{2} - 4 x_{1} = 4 (x_{2} - x_{1})$