Question

Considering that binding energy per nucleon of the deuterium isotope's nucleus $\left({}_1^{2}H\right)is1.12MeV$ ,what will be the same for the protium isotope's nucleus $\left({}_1^{1}H\right)$

• A. 1.12 MeV
• B. 0.56 MeV
• C. 0.28 MeV
• D. 0 eV

Answer 1

The correct option is D 0 eV

Protium, or the common hydrogen isotope $\left({}_1^{1}H\right)$ that we know, has only one nucleon in the nucleus - a proton. There are no other neutrons that this proton needs to sit in a bound state with.

Therefore its 'binding' energy is said to be zero. Which means protium's binding energy per nucleon is $0eV$ as well