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Calculate The binding energy per nucl...

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The binding energy per nucleon of `._(6)^(12)C` nucleus. Nuclear mass of `._(6)^(12)C=12.000000 u`, mass of proton = 1.007825 u and mass of neutron = 1.008665 u.

A

2.675 MeV

B

7.675 MeV

C

0 MeV

D

3.675 MeV

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The correct Answer is:
B
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The binding energy per nucleon of ""_(7)N^(14) nucleus is: (Mass of ""_(7)N^(14) = 14.00307 u ) mass of proton = 1.007825 u mass of neutron = 1.008665 u

Calculate the binding energy per nucleon of ._26Fe^(56) nucleus. Given that mass of ._26Fe^(56)=55.934939u , mass of proton =1.007825u and mass of neutron =1.008665 u and 1u=931MeV .

Calculate the binding energy per nucleon of ._(20)^(40)Ca . Given that mass of ._(20)^(40)Ca nucleus = 39.962589 u , mass of proton = 1.007825 u . Mass of Neutron = 1.008665 u and 1 u is equivalent to 931 MeV .

What is the binding energy per nucleon of _(6)C^(12) nucleus? Given , mass of C^(12) (m_(c))_(m) = 12.000 u Mass of proton m_(p) = 1.0078 u Mass of neutron m_(n) = 1.0087 u and 1 amu = 931.4 MeV

Calculate the binding energy per nucleon of ._(20)Ca^(40) nucleus. Mass of (._(20)Ca^(40)) = 39.962591 am u .

Calculate the binding energy for nucleon of . _6^(12) C nucleus, if mass of proton m_p = 10078 u , mass of neutron m_n = 1.0087 u , mass of C_12, m_C =12.0000 u , and 1 u =931 .4 MeV .

Calculate the average binding energy per nucleon of ._(41)^(93)Nb having mass 9.2.906 u..

AAKASH INSTITUTE-NUCLEI-ASSIGNMENT (SECTION-A)
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  2. In a nuclear fusion reaction, if the energy is released then

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  8. The atomic mass of .(7)N^(15) is 15.000108 "amu" and that of .(8)O^(16...

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