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Nuclear Fusion

The B.E/A r...

Question

The $\frac{B . E}{A}$ ratio for deuteron and $α$ - particle are $X_{1}$ and $X_{2}$ respectively. The energy released in the fusion of deuteron into $α$ -particle is

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

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2 (X_{2} - X_{1})

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

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\frac{X_{2} - X 1}{4}

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Solution

The correct option is A $4 (X_{2} - X_{1})$
Energy released is given as per the reaction
$_{1} H^{2} +_{1} H^{2} =_{2} H e^{4} + Q$
In the two deuterons $(_{1} H^{2})$ there are total 4 nucleons and in the alpha particle $(_{2} H e^{4})$ there are 4 nucleons.
Thus we get
$Q = 4 (X_{2} - X_{1})$

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Similar questions

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 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,

Q. 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

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}

Q. The following system of equations

x_{1} + x_{2} + 2 x_{3} = 1, x_{1} + 2 x_{2} + 3 x_{3} = 2, x_{1} + 4 x_{2} + α x_{3} = 4

has a unique solution. The only possible value for

α

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The B.EA ratio for deuteron and α- particle are X1 and X2respectively. The energy released in the fusion of deuteron into α-particle is

The correct option is A 4(X2−X1)Energy released is given as per the reaction1H2+1H2=2He4+QIn the two deuterons (1H2) there are total 4 nucleons and in the alpha particle (2He4) there are 4 nucleons.Thus we getQ=4(X2−X1)

The $\frac{B . E}{A}$ ratio for deuteron and $α$ - particle are $X_{1}$ and $X_{2}$ respectively. The energy released in the fusion of deuteron into $α$ -particle is

The correct option is A $4 (X_{2} - X_{1})$
Energy released is given as per the reaction
$_{1} H^{2} +_{1} H^{2} =_{2} H e^{4} + Q$
In the two deuterons $(_{1} H^{2})$ there are total 4 nucleons and in the alpha particle $(_{2} H e^{4})$ there are 4 nucleons.
Thus we get
$Q = 4 (X_{2} - X_{1})$