128.The binding energy of alpha is 7.047 MeV per nucleon and the binding energy of deuteron is 1.112MeV per nucleon. Then in the fusion reaction given, the Q released is

Open in App

Solution

Suggest Corrections

Similar questions

The binding energy of deuteron $(_{1}^{2} H)$ is 1.15 MeV per nucleon and an alpha particle $(_{2}^{4} H e)$ has a binding energy of 7.1 MeV per nucleon. Then in the reaction
$_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e + Q$
the energy Q released is

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

)

The binding energy of deuteron $(_{1}^{2} H)$ is 1.15 MeV per nucleon and an alpha particle $(_{2}^{4} H e)$ has a binding energy of 7.1 MeV per nucleon. Then in the reaction

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

the energy Q released is approximately

Q. Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron

= 1.1 M e V

and for helium

= 7.0 M e V

)

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

Join BYJU'S Learning Program