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

23.6 M e V

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

25.2 M e V

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

16.6 M e V

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

13.6 M e V

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

Open in App

Solution

The correct option is A $23.6 M e V$
The total binding energy of $_{1}^{2} H$ = $2 \times 1.1$ = $2.2 M e V$
There are two such atoms hence
total binding energy before reaction = $2 \times 2.2 = 4.4 M e V$
Total binding energy after reaction = $4 \times$ binding energy per nucleon of $_{2}^{4} H e$
$= 4 \times 7 = 28 M e V$
Hence, energy released = $28 - 4.4 = 23.6 M e V$

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. Binding energy per nucleon of

_{1} H^{2}

and

_{2} H e^{4}

are 1.1 MeV and 7.0 MeV respectively. Energy released in the process

_{1} H^{2} +_{1} H^{2} =_{2} H e^{4}

Q. The binding energy per nucleon of deuteron

(_{1}^{2} H)

and helium nucleus

(_{2}^{4} H e)

is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

Q. Binding energy per nucleon of

_{1} H^{2}

and

_{2} H e^{4}

are

1.1

MeV and

7.0

MeV respectively. Energy released in the process

_{1} H^{2} +_{1} H^{2} ⟶_{2} H e^{4}

is -

Q. The binding energy per nucleon of deuteron

(_{1}^{2} H)

and helium nucleus

(_{2}^{4} H e)

is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

Join BYJU'S Learning Program

The binding energy per nucleon of 21H and 42He are 1.1 MeV and 7.0 MeV respectively. Energy released in the process 21H+21H→42He is

The correct option is A 23.6 MeVThe total binding energy of 21H = 2×1.1 =2.2 MeV There are two such atoms hence total binding energy before reaction = 2×2.2=4.4 MeV Total binding energy after reaction =4× binding energy per nucleon of 42He =4×7=28 MeV Hence, energy released = 28−4.4=23.6 MeV

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