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Mechanism of Nuclear Fission

The binding e...

Question

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

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 $4 (x_{2} - x_{1})$
In the given reaction,

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

Binding energy of reactants is, $E_{b 1} = 2 x_{1} + 2 x_{1} = 4 x_{1}$

And, binding energy of the product is,
$E_{b 2} = 4 x_{2}$

The energy released in the reaction is,
$E_{b 2} - E_{b 1} = 4 x_{2} - 4 x_{1} = 4 (x_{2} - x_{1})$

Hence, $(B)$ is the correct answer.

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Similar questions

Q. The binding energy per nucleon for a

deuteron

and an

α - particle

x_{1}

and

x_{2}

respectively. The energy

Q

released in the reaction

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

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:

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

Q. Binding energy per nucleon of

_{1} H^{2}

and

_{2} H e^{4}

are

1.1 M e V

and

7.0 M e V

, respectively. Energy released in the process

_{1} H^{2} +_{1} H^{2} ⟶_{2} H e^{4}

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

, the energy

Q

released is
(take mass of

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

;
mass of

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

)

Q. Binding energy per nucleon of

_{1} H^{2}

and

_{2} H e^{4}

are 1.1 MeV and 7.0 MeV respectively. Energy released in the process

_{1} H^{2} +_{1} H^{2} =_{2} H e^{4}

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The binding energy per nucleon for a deuteron and an alpha particle, are x1 and x2 respectively. What will be the energy released in the following reaction? 1H2+ 1H2→ 2He4

The correct option is B 4(x2−x1)In the given reaction, 1H2+ 1H2→ 2He4 Binding energy of reactants is, Eb1=2x1+2x1=4x1 And, binding energy of the product is, Eb2=4x2 The energy released in the reaction is, Eb2−Eb1=4x2−4x1=4(x2−x1) Hence, (B) is the correct answer.

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