Question

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

• A. $0.0048$
• B. $0.0024$
• C. $0.0012$
• D. $0.0006$

Answer 1

Answer: (B)

$0.0024$

Answer 2

: Binding energy per nucleon is

$E_{bn} = \frac{E_{b}}{A}$

For deuterium $A = 2$

$\therefore 1.115MeV = \frac{E_{b}}{2}$

or$E_{b} = 2 \times 1.115MeVorE_{b} = \Delta Mc^{2}$

mass defect,

$\Delta M = \frac{2 \times 1.115}{931.5}u\lbrack\because 1u = 931.5MeV/c^{2}\rbrack$

$= 0.0024u$