Question

if the binding energy performance nucleon of deuterium is 1.115MeV,it's mass defect in atomic mass unit is​

Answer 1

Answer:

H

2

+

1

H

2

→

2

He

4

+ΔE

The binding energy per nucleon of a deuteron =1.1MeV

∴ Total binding energy = 2×1.1 =2.2MeV

The binding energy per nucleon of a helium nuclei =7MeV

∴ Total binding energy = 4×7=28MeV

∴ Hence, energy released

ΔE=(28−2×2.2) =23.6MeV

Explanation:

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Answer 2

Answer:

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