Question 145: Atoms & Nuclei

Question

Energy released when two deuterons $\left({ }_1 \mathrm{H}^2\right)$ fuse to form a helium nucleus $\left({ }_2 \mathrm{He}^4\right)$ is :
(Given : Binding energy per nucleon of ${ }_1 \mathrm{H}^2=1.1 \mathrm{MeV}$ and binding energy per nucleon of ${ }_2 \mathrm{He}^4=7.0 \mathrm{MeV}$ )

Accepted Answer

To calculate the energy released when two deuterons (${}_1 \mathrm{H}^2$) fuse to form a helium nucleus (${}_2 \mathrm{He}^4$), consider the following:
The reaction is represented as:
$$ {}_1{H^2} + {}_1{H^2} 	o {}_2{He^4} $$
Given the binding energy per nucleon:
For ${}_1 \mathrm{H}^2$: 1.1 MeV
For ${}_2 \mathrm{He}^4$: 7.0 MeV
Calculating Binding Energy
Binding Energy for Reactants
:
Each deuteron (${}_1 \mathrm{H}^2$) has a binding energy of $1.1 \, 	ext{MeV}$ per nucleon.
Since there are two nucleons in a deuteron, the total binding energy for one deuteron is $1.1 	imes 2 = 2.2 \, 	ext{MeV}$.
Therefore, for two deuterons: $2.2 	imes 2 = 4.4 \, 	ext{MeV}$.
Binding Energy for Product
:
For ${}_2 \mathrm{He}^4$: Each of the four nucleons has a binding energy of $7.0 \, 	ext{MeV}$.
Total binding energy for the helium nucleus: $7.0 	imes 4 = 28.0 \, 	ext{MeV}$.
Energy Released (Q)
The energy released, $ \mathrm{Q} $, is the difference between the binding energy of the products and the reactants:
$ \mathrm{Q} = \mathrm{BE}_{	ext{product}} - \mathrm{BE}_{	ext{reactants}} = 28.0 \, 	ext{MeV} - 4.4 \, 	ext{MeV} = 23.6 \, 	ext{MeV} $
Thus, the energy released when two deuterons fuse to form a helium nucleus is
23.6 MeV
.

Answer

26.8 MeV

Answer

8.1 MeV

Answer

23.6 MeV

Answer

5.9 MeV

Atoms & Nuclei

Energy released when two deuterons \left({ }_1 \mathrm{H}^2\right) fuse to form a helium nucleus \left({ }_2 \mathrm{He}^4\right) is : (Given : Binding energy per nucleon of { }_1 \mathrm{H}^2=1.1 \mathrm{MeV} and binding energy per nucleon of { }_2 \mathrm{He}^4=7.0 \mathrm{MeV} )

Choose the correct answer: