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If the binding energy of the deuterium i...

If the binding energy of the deuterium is `2.23 MeV`. The mass defect given in a.m.u. is.

A

0.0024

B

0.0012

C

`-0.0012`

D

`-0.0024`

Text Solution

Verified by Experts

The correct Answer is:
A
B.E. `=(Deltam) xx 931` MeV
So, `Deltam` must be in amu.
`2.23 =Deltam xx 931`
`Deltam = 2.395 xx 10^(-3)`
`=0.002395 u = +0.0024`u
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