The binding energy per nucleon for a deuteron and an -particle is x1 and x2 respectively- The energy Q released in the reaction 1H2-1H2- 2He4 -Q- is

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Mechanism of Nuclear Fission

The binding e...

Question

The binding energy per nucleon for a $deuteron$ and an $α - particle$ is $x_{1}$ and $x_{2}$ respectively. The energy $Q$ released in the reaction $_{1} H^{2} +_{1} H^{2} \to_{2} {He}^{4} + Q,$ is

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

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

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

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

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Solution

The correct option is B $4 (x_{2} - x_{1})$
Binding Energy of a deuteron,

$B . E of_{1} H^{2} = 2 x_{1}$

Binding Energy of a helium,

$B . E of_{2} {He}^{4} = 4 x_{2}$

By conservation of energy

$Q = (B . E)_{product} - (B . E)_{reactants}$

$\Rightarrow Q = B . E of_{2} {He}^{4} - 2 (B . E of_{1} H^{2})$

$\Rightarrow Q = 4 x_{2} - 2 (2 x_{1})$

$\Rightarrow Q = 4 (x_{2} - x_{1})$

Hence, option $(B)$ is correct.

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Similar questions

Q. 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}

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. 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 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. 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}

is -

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The binding energy per nucleon for a deuteron and an α−particle is x1 and x2 respectively. The energy Q released in the reaction 1H2+1H2→2He4+Q, is

The correct option is B 4(x2−x1)Binding Energy of a deuteron, B.E of 1H2=2x1 Binding Energy of a helium, B.E of 2He4=4x2 By conservation of energy Q=(B.E)product−(B.E)reactants ⇒Q=B.E of 2He4−2(B.E of 1H2) ⇒Q=4x2−2(2x1) ⇒Q=4(x2−x1) Hence, option (B) is correct.

The binding energy per nucleon for a $deuteron$ and an $α - particle$ is $x_{1}$ and $x_{2}$ respectively. The energy $Q$ released in the reaction $_{1} H^{2} +_{1} H^{2} \to_{2} {He}^{4} + Q,$ is