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., ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

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,

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:

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

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

Q. 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

Join BYJU'S Learning Program

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