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If the mass of proton= 1.008 a.m.u. and ...

If the mass of proton= 1.008 a.m.u. and mass of neutron=1.009a.m.u. then binding energy per nucleon for `._(4)Be^9` (mass=9.012 amu) would be-

A

`0.06772 MeV`

B

`0.672 MeV`

C

`6.724 MeV`

D

`67.2 MeV`

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

  • The mass of proton is 1.0073 u and that of neutron is 1.0087 u ( u= atomic mass unit). The binding energy of ._(2)He^(4) is (mass of helium nucleus =4.0015 u )

    A
    `28.4` MeV
    B
    `0.061 u`
    C
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    D
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  • If the mass defect of ""_(4)^(9)X is 0.09 a.m.u. then the binding energy per nucleon is (1 a.m.u= 931.5 MeV)

    A
    9.315 MeV
    B
    931.5 MeV
    C
    83.8 MeV
    D
    8.38 MeV

  • One a.m.u. has the mass equal to

    A
    `1.66xx10^(-22) g`
    B
    `1.66xx10^(-24)g`
    C
    1 g
    D
    `(1)/(12)g`

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