The binding energy of deuteron 21H is 1-15 MeV per nucleon and an alpha particle 42He has a binding energy of 7-1 MeV per nucleon- Then in the reaction21H - 21H - 42 He - Qthe energy Q released is

Binding energy of ^2_1 H = 1.15 × number of nucelons = 1.15 × 2 = 2.3 MeV. Total binding energy of reactants = 2.3 + 2.3 = 4.6 MeV. Binding energy of ^4_2He = 7 ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Binding Energy

The binding e...

Question

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

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. 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. 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.
Hence the correct choice is (c)

Suggest Corrections

Similar questions

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

Q. The binding energies per nucleon for deuteron .(

_{1}^{2} H

) and helium (

_{2}^{4} H e

) are 1.1 MeV and 7.0 MeV respectively. The energy released, when two deuterons fuse to form a helium nucleus is

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

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

_{3}^{7} Li

and

_{2}^{4} He

nuclei are

5.60 MeV

and

7.06 MeV

,respectively. In the nuclear reaction

_{3}^{7} Li +_{1}^{1} H \to_{2}^{4} He + Q,

the value of energy

Q

released is

Join BYJU'S Learning Program

The binding energy of deuteron (21H) is 1.15 MeV per nucleon and an alpha particle (42He) has a binding energy of 7.1 MeV per nucleon. Then in the reaction 21H+21H→42He+Q the energy Q released is

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