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Question

Chemistry Question on Nuclear physics

The mass of helium atom of mass number 44 is 4.00264.0026 amu, while that of the neutron and proton are 1.00871.0087 and 1.00781.0078 respectively on the same scale. Hence, the nuclear binding energy per nucleon in the helium atom is nearly

A

5 MeV

B

7 MeV

C

10 MeV

D

14 MeV

Answer

7 MeV

Explanation

Solution

He atom has 2p+2n2p+2n
Hence,
Δm=(2×1.0078+2×1.0087)4.0026\Delta m=(2\times 1.0078+2\times 1.0087)-4.0026
=0.0304amu=0.0304\,amu
\therefore energy released
=0.0304×931.5MeV=0.0304\times 931.5\,MeV
=28.3MeV=28.3\,MeV
Binding energy per nucleon
=28.34=7MeV=\frac{28.3}{4}=7\,MeV (approximately)