Question

The mass defect of the nucleus of helium is 0.0303 amu .The binding energy per nucleon in Mev is nearly

Answer 1

Answer:-

Mass defect of helium nucleus ∆M$=0.0303\:amu$

Binding energy E =∆MC²

=∆$M ×931.5\:MeV$

$⟹E= 0.0303 × 931.5$

$28\:Mev$

Number of nucleon in helium nucleus $=4$

Thus binding energy per nucleon $=\frac{28}{4}$

$=7\:Mev$