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If the binding energy per nucleon of deu...

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

Text Solution

Verified by Experts

The correct Answer is:
B
Binding energy per nucleon in `E_(bn)=(E_(b))/(A)`
For deuteron, `A=2`
`therefore" 1.115 MeV"=(E_(b))/(2)`
`E_(b)=2xx1.115MeV`
`E_(b)=DeltaMc^(2)`
Mass defect, `DeltaM=(2xx1.115)/(931.5)u=0.0024u`
`" "[because 1u="931.5 MeV/c"^(2)]`
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