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The binding energy per nucleon of deuter...

The binding energy per nucleon of deuterium and helium atom is `1.1 MeV` and `7.0 MeV`. If two deuterium nuclei fuse to form helium atom, the energy released is.

A

23.6MeV

B

25.8 MeV

C

30.2 MeV

D

32.4 MeV

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The correct Answer is:
To find the energy released when two deuterium nuclei fuse to form a helium atom, we can follow these steps: ### Step 1: Understand the reaction We start with two deuterium nuclei (each with atomic number 1 and atomic mass 2) and they fuse to form one helium nucleus (with atomic number 2 and atomic mass 4). The reaction can be represented as: \[ 2 \, ^2_1H \rightarrow \, ^4_2He + \text{Energy} \] ### Step 2: Calculate the total binding energy of the reactants (deuterium) The binding energy per nucleon for deuterium is given as 1.1 MeV. Since each deuterium nucleus has 2 nucleons, the total binding energy for one deuterium nucleus is: \[ \text{Total binding energy for one deuterium} = 2 \times 1.1 \, \text{MeV} = 2.2 \, \text{MeV} \] Since we have two deuterium nuclei, the total binding energy for both deuterium nuclei is: \[ \text{Total binding energy for two deuterium nuclei} = 2 \times 2.2 \, \text{MeV} = 2.2 \, \text{MeV} \] ### Step 3: Calculate the total binding energy of the product (helium) The binding energy per nucleon for helium is given as 7.0 MeV. Since the helium nucleus has 4 nucleons, the total binding energy for helium is: \[ \text{Total binding energy for helium} = 4 \times 7.0 \, \text{MeV} = 28.0 \, \text{MeV} \] ### Step 4: Calculate the energy released in the fusion reaction The energy released during the fusion reaction can be calculated by finding the difference between the total binding energy of the product (helium) and the total binding energy of the reactants (deuterium): \[ \text{Energy released} = \text{Total binding energy of helium} - \text{Total binding energy of deuterium} \] \[ \text{Energy released} = 28.0 \, \text{MeV} - 2.2 \, \text{MeV} = 25.8 \, \text{MeV} \] ### Final Answer The energy released when two deuterium nuclei fuse to form a helium atom is **25.8 MeV**. ---
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