If the binding energy per nucleon of deuterium is 1.115 MeV, its mass defect in atomic mass unit is

Question

If the binding energy per nucleon of deuterium is 1.115 MeV, its mass defect in atomic mass unit is (A) 0.0048 (B) 0.0024 (C) 0.0012 (D) 0.0006

Tardigrade Admin · Accepted Answer

Binding energy per nucleon is Ebn=(Eb/A) For deuterium, A = 2 ∴ 1.115 MeV =(Eb/2) or Eb=2×1.115 MeV or Eb=Δ Mc2 Mass defect, Δ M=(2×1.115/931.5)u [∵1 u=931.5 MeV/ c2] =0.0024 u

Q. If the binding energy per nucleon of deuterium is $1.115 M e V$ , its mass defect in atomic mass unit is

$0.0048$

$0.0024$

$0.0012$

$0.0006$

Binding energy per nucleon is
$E_{bn} = \frac{E _{b}}{A}$
For deuterium, $A = 2$
$∴ 1.115 M e V = \frac{E _{b}}{2}$
or $E_{b} = 2 \times 1.115 M e V$ or $E_{b} = Δ M c^{2}$
Mass defect,
$Δ M = \frac{2 \times 1.115}{931.5} u [∵ 1 u = 931.5 M e V / c^{2}]$
$= 0.0024 u$

Q. If the binding energy per nucleon of deuterium is 1.115MeV, its mass defect in atomic mass unit is

0.0048

0.0024

0.0012

0.0006