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The mass of proton is 1.0073 u and that ...

The mass of proton is 1.0073 u and that of neutron is 1.0087 u ( u=atomic mass unit ). The binding energy of `._2^4He`, if mass of `._2^4He ` is `4.0015 u`, is

A

0.0305 erg

B

0.0305 J

C

28.4 MeV

D

0.061 u

Text Solution

Verified by Experts

The correct Answer is:
C
`Deltam=2m_p + 2m_n -m(._2^4He)`
=2 x 1.0073 + 2 x 1.0087 - 4.0015 =0.0305 u
Binding energy = 0.0305 x 931 MeV = 28.4 MeV
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Mass of Helium nucleus = 4 . 00015 amu and mass of a proton =1 .0073 amu mass of neutron= 1.0087 amu Calculate the mass defect and energy liberated in formation of He nucleus and also evaluate binding energy of ._(2)He^(4) nucleus.

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
    `0.0305 J`
    D
    `0.0305` erg

  • One atomic mass unit (u) is equivalent to an energy of :

    A
    931 eV
    B
    9.31 MeV
    C
    1 MeV
    D
    931 MeV

  • 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
    0.065 MeV
    B
    60 .44MeV
    C
    0.7 MeV
    D
    6.72 MeV

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