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If the mass defect of .(5)B^(11) is 0.08...

If the mass defect of `._(5)B^(11)` is 0.081 u, its average binding energy (in MeV) is

A

`8.60`

B

`6.85`

C

`5.60`

D

`5.86`

Text Solution

Verified by Experts

The correct Answer is:
b
Given , `Delta` m for `._(5)B^(11) = 0.081 u`
Number of nucleons `= 11`
Binding energy `= 931xxDelta` m MeV
`= 931xx0.081 = 75.411 MeV`
`"Average binding energy" = ("Binding energy")/("Number of nucleons")`
`= (75.411)/(11)`
`= 6.85 MeV`
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