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

1 MeV

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

11.9 MeV

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

23.8 MeV

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

931 MeV

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

Open in App

Solution

The correct option is C
23.8 MeV

Binding energy of $_{1}^{2} H = 1.15 \times$ number of nucelons = $1.15 \times 2 = 2.3$ MeV.

Total binding energy of reactants = 2.3 + 2.3 = 4.6 MeV.

Binding energy of $_{2}^{4} H e = 7.1 \times$ number of nucleons = $7.1 \times 4 = 28.4$ MeV.

Therefore , Q= 28.4 - 4.6 = 23.8 MeV.

Suggest Corrections

Similar questions

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 binding energy per nucleon of

_{1}^{2} H

and

_{2}^{4} H e

are

1.1 M e V

and

7.0 M e V

respectively. Energy released in the process

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

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