Solved: PHYSICS Question for NEET | Sushrut

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NEET PHYSICSEasy

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 a helium atom, the energy released is:

A

19.2 MeV

B

23.6 MeV

C

26.9 MeV

D

13.9 MeV

Step-by-Step Solution

Identify Nucleons:

Deuterium ( ${}_{1}^{2}\text{H}$ ) has a mass number $A=2$ (1 proton, 1 neutron).
Helium ( ${}_{2}^{4}\text{He}$ ) has a mass number $A=4$ (2 protons, 2 neutrons).

Calculate Initial Binding Energy (Reactants):

Binding Energy per nucleon for Deuterium = $1.1 \text{ MeV}$ .
Total Binding Energy for one Deuterium nucleus = $2 \times 1.1 = 2.2 \text{ MeV}$ .
Since two Deuterium nuclei fuse, total initial Binding Energy = $2 \times 2.2 = 4.4 \text{ MeV}$ .

Calculate Final Binding Energy (Products):

Binding Energy per nucleon for Helium = $7.0 \text{ MeV}$ .
Total Binding Energy for the Helium nucleus = $4 \times 7.0 = 28.0 \text{ MeV}$ .

Calculate Energy Released ( $Q$ ):

The energy released in a nuclear reaction is the difference between the total binding energy of the products and the reactants .
$Q = (BE)_{\text{products}} - (BE)_{\text{reactants}}$
$Q = 28.0 \text{ MeV} - 4.4 \text{ MeV} = 23.6 \text{ MeV}$ .

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