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The value of binding energy per nucleon ...

The value of binding energy per nucleon of `._(20)^(40)Ca` nucleus is Given :
Mass of `._(20)^(40)Ca` nucleus `=39.962589u`
`"Mass of proton "=1.007825u`
`"Mass of neutron "=1.008665u`
and `1u="931 MeV C"^(-2)`

A

4.55 Mev

B

8.55 Mev

C

6.55 Mev

D

7.55 Mev

Text Solution

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The correct Answer is:
B
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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 .

Calculate the binding energy per nucleon of ._17^35Cl nucleus. Given that mass of ._17^35Cl nucleus = 34.98000 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 mass defect and binding energy per nucleon of ""_(10)^(20)Ne , given Mass of ""_(10)^(20)Ne = 19.992397 u Mass of ""_1^(1)H = 1.007825 u Mass of on = ""_(0)^(1)1.008665 u

What is the binding energy per nucleon in ._2He^4 Given , Mass of ._2He^4 =4.002604 amu Mass of proton =1.007825 amu Mass of neutron =1.008665 amu

What is the binding energy per nucleon in ._2He^4 Given , Mass of ._2He^4 =4.002604 amu Mass of proton =1.007825 amu Mass of neutron =1.008665 amu

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The energy required to separate the typical middle mass nucleus ._(50)^(120)Sn into its constituent nucleons ( Mass of ._(50)^(120)sn=119.902199u , mass of proton =1.007825 u and mass of neutron =1.008665 u )

Binding energy per nucleon of ._50Sn^120 approximately will be. [ Atomic mass of ._50Sn^120 is 120.500 u and that of ._1H^1 is 1.007u . mass of neutron = 1.008u , 1u= 931 Mev ]

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